It is easily shown that f(2n(3n+1)) = 6n+1, n≥0, and f(2n(3n-1)) = 6n-1, n≥1/3.

Clearly then f(2n(3n+1)) = 6n+1 and f(2n(3n-1)) = 6n-1, n∈ω.

The set of primes ≥ 5 ⊂ {6n+1: n∈ω} ∪ {6n-1: n∈ω} and so f() has an infinite number of ‘integer’ points including **all** of those where the function values are primes ≥ 5.

E.g.

n=1 and so f(8)=7 and f(4)=5

n=2 and so f(28)=13 and f(20)=11

n=3 and so f(60)=19 and f(48)=17

etc.

Clearly some of these integer points have composite function values.

E.g when n=4 then f(104)=25=5² and when n=8, f(400)=49=7² .

**The case (6n+1)**

If we let g(n)=2n(3n+1), n∈ω, then f(g(n))=6n+1. Also, it is easily shown that

f(g(g(n)))=fogog(n)=fg²(n)=(6n+1)² , i.e. composite.

Again, fg³(n)=((6n+1)²)²=(6n+1)^4. This extends to fg^(a+1)(n)=(6n+1)^(2^a).

**The case (6n-1)**

If we let g(n)=2n(3n-1), n∈ω, then f(g(n))=6n-1. Also, if h(n)=6n, n∈ω, then

f(g(h(n))) = (6n)² – 1 = (6n+1)(6n-1), i.e. composite.

If we let P(5) = {prime numbers ≥ 5}, A = {6n+1: n∈ω} and B = {6n-1: n∈ω} then it is easy to show that P(5) ⊂ A ∪ B, A ∩ B = Ø, and (A, .) is a commutative semi-group where . is ordinary multiplication. Also, if a∈A and b∈B then it is clear that a²∈A, b²∈A and ab∈B.

Since P(5) is infinite and P(5) ⊂ A ∪ B, then A and B cannot both contain at most a finite number of primes, so A, B, or both, must contain an infinite number of primes.

**The set B contains an infinite number of primes.**

If P_{B} is the set of primes in B and C_{B} is the set of composites in B then B=P_{B} ∪ C_{B} and obviously

P_{B} ∩ C_{B} = ∅.

Suppose now that P_{B} is finite. We will try to show that this leads to a contradiction.

P_{B} is finite means that P_{B} can be listed as a set of **n** primes in order, i.e.

P_{B} = {p_{1}, p_{2}, … , p_{n}} where p_{1}<p_{2} … <p_{n}.

Now consider x = 6(∏_{i=1 .. n} p_{i)} – 1. Clearly x∈B. If x∈P_{B} then x≤p_{n} which is clearly not true and so x∈C_{B} .

If all of the primes dividing x belong to A, then x∈A which is not possible since A ∩ B = Ø, and so at least one prime dividing x belongs to B. Since x = 6(∏_{i=1 .. n} p_{i)} – 1 that means some prime in B divides into 1, i.e. impossible.

Hence the result is proved.

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It is straightforward to show that if **p** is a prime number ≥ 5 then there is a positive integer **d** (say) such that either p = 6d+1 or p = 6d – 1. Hence, (p – 6d)² = 1.

Multiplying out the left-hand side of (p – 6d)² = 1 gives p² – 12dp + 36d² = 1.

It then follows easily that p² – 1 = 12d(p – 3d). If d is even then p² – 1 must be a multiple of 24 and, if d is odd, p – 3d is also even and so again 24|p² – 1.

Now the equation p² – 12dp + 36d² = 1 can be rearranged to give p² + 36d² = 1 + 12dp, which is suggestive of the form of a Pythagorean triple on the left-hand side of p² + 36d² = 1 + 12dp.

We will look for a Pythagorean triple of the form p² + (λd)² = (1 +λd)² where λ should at least be a positive integer. Multiplying out p² + (λd)² = (1 +λd)² gives p² = 1 + 2λd.

However, p² = 1 + 12dp – 36d², and so 2λd = 12dp – 36d², or λ = 6p – 18d = 6(p – 3d).

Since p = 6d + 1 or 6d – 1 for some d∈ω, it follows that λ must be a positive integer. It is also clear that if p is a prime ≥ 5 then d=int((p+1)/6).

So, we have that p² + (6(p – 3d)d)² = (6(p – 3d)d + 1)² or, that (p, 6(p – 3d)d, 6(p – 3d)d+1) is a Pythagorean triple. This triple must also be primitive.

There is a chance also that 6(p – 3d)d+1 might also be prime, and so we may discover a ‘two-prime’ ppt. For example if we start with p=61, then using d=int((p+1)/6), we calculate d=10 giving (61, 1860, 1861) where it is easily checked that 1861 is also prime.

Again if p=131, then d=22 and so we have (131, 8580, 8581), where 8581 is also prime.

**Note: **(p, 6(p – 3d)d, 6(p – 3d)d+1) is the same ppt as that obtained from

(2n+1, 2n² + 2n, 2n² + 2n + 1) by letting p = 2n + 1 and then using p² – 1 = 12d(p – 3d).

If we wish to look for examples of two-prime ppt’s we might consider

(6d+1, 6d(3d+1), 6d(3d+1)+1) and (6d-1, 6d(3d-1), 6d(3d-1)+1),

i.e. letting p=6d+1 or 6d-1 in (p, 6(p – 3d)d, 6(p – 3d)d+1) and allowing d to be

any integer ≥ 1.

We can easily rule out the ‘x’ and ‘z’ terms that are multiples of 5 and so we arrive at

(6d+1, 6d(3d+1), 6d(3d+1)+1) can only have d=5n-2 or 5n where n is any positive integer, and

(6d-1, 6d(3d-1), 6d(3d-1)+1) can only have d=5n-3 or 5n, where n is any positive integer.

If we allow d to range through the values 1, 2, … to 200 we can generate the following

‘two-prime ppt’s’ from (6d+1, 6d(3d+1), 6d(3d+1)+1) …

( 19 , 180 , 181 )

( 61 , 1860 , 1861 )

( 79 , 3120 , 3121 )

( 139 , 9660 , 9661 )

( 181 , 16380 , 16381 )

( 199 , 19800 , 19801 )

( 271 , 36720 , 36721 )

( 349 , 60900 , 60901 )

( 379 , 71820 , 71821 )

( 409 , 83640 , 83641 )

( 571 , 163020 , 163021 )

( 631 , 199080 , 199081 )

( 661 , 218460 , 218461 )

( 739 , 273060 , 273061 )

( 751 , 282000 , 282001 )

( 991 , 491040 , 491041 )

( 1039 , 539760 , 539761 )

( 1051 , 552300 , 552301 )

( 1069 , 571380 , 571381 )

( 1129 , 637320 , 637321 )

( 1171 , 685620 , 685621 )

and from (6d-1, 6d(3d-1), 6d(3d-1)+1) …

( 5 , 12 , 13 )

( 11 , 60 , 61 )

( 29 , 420 , 421 )

( 59 , 1740 , 1741 )

( 71 , 2520 , 2521 )

( 101 , 5100 , 5101 )

( 131 , 8580 , 8581 )

( 449 , 100800 , 100801 )

( 461 , 106260 , 106261 )

( 521 , 135720 , 135721 )

( 569 , 161880 , 161881 )

( 641 , 205440 , 205441 )

( 821 , 337020 , 337021 )

( 881 , 388080 , 388081 )

( 929 , 431520 , 431521 )

( 1031 , 531480 , 531481 )

( 1091 , 595140 , 595141 )

( 1151 , 662400 , 662401 )

( 1181 , 697380 , 697381 )

After a very brief look at the formulae that generate infinite sequences of PPT’s it seems that

(6d+1, 6d(3d+1), 6d(3d+1)+1) and (6d-1, 6d(3d-1), 6d(3d-1)+1), d∈ω, may be the only two such formulae that have that property. That remains to be proved or disproved.

If p and q are odd primes ≥ 5 and q>p then it is clearly true that 24| (q² – p²) and so if q² – p² is also a square then q² – p² = (12r)² for some r∈ω.

**Euclid’s Formulae and Primitive Pythagorean Triples (PPT’s).**

These formulae allow PPT’s to be generated using 2 (positive integer) parameters, usually denoted by m and n. The PPT’s (usually) take the form ( m^{2} – n^{2}, 2mn, m^{2}+n^{2}) where m>n≥1, m and n are coprime and m-n is odd. Clearly, m^{2}+n^{2} is the largest number in each triple. It is easier to spot patterns in generated triples if these are ordered as in (x, y, z) where x<y<z.

Hence there are 2 cases that can occur, i.e. (m^{2}– n^{2}, 2mn, m^{2}+n^{2}) where m^{2}-n^{2} < 2mn and

(2mn, m^{2}-n^{2},m^{2}+n^{2}) where 2mn < m^{2}-n^{2}.

If we look at the case (2mn, m^{2}-n^{2}, m^{2}+n^{2}) where 2mn < m² – n² and assume that m² – n² and m² + n² are both prime we see that this cannot occur. Since m² – n² factorises into (m+n)(m-n) the smaller factor must be 1 i.e. m-n=1. So, (2mn, m^{2}-n^{2}, m^{2}+n^{2}) becomes

(2(n+1)n, 2n+1, (n+1)² + n²). It follows then that 2(n+1)n < 2n+1 or 2n² < 1 or that n≤0, which is not possible since n≥1.

Hence, two prime PPT’s must take the form (m^{2}– n^{2}, 2mn, m^{2}+n^{2}) where m^{2}-n^{2}<2mn. A repeat of the argument above shows that m=n+1 and so all two-prime PPT’s must take the form (2n+1, 2n²+2n, 2n²+2n +1) where n∈ω.

For a two-prime PPT we must also have that 12|2n²+2n, i.e. 12|2n(n+1), and so it must be true that 3|n or 3|(n+1).

If 3|n then n=3m, m∈ω, and so (2n+1, 2n²+2n, 2n²+2n +1) becomes …

(6m+1, 6m(3m+1), 6m(3m+1)+1), while if 3|(n+1) then n=3m-1, in which case

(2n+1, 2n²+2n, 2n²+2n +1) becomes (6m-1, 6m(3m-1), 6m(3m-1)+1).

**Hence the result is proved.**

It is easily proved that for any PPT (x,y,z) (say), exactly one of x, y and z is divisible by 5. So, if (x,y,z) is also two-prime, and x<y<z, then, if 5|x, (x,y,z) = (5,12,13) and, if 5|z, then (x,y,z) = (3,4,5). Otherwise, 5|y. All other two-prime PPT’s take the form (p, 60r, q) where

p<60r<q and p>5, p and q prime.

Such two-prime PPT’s can be generated by …

(6m+1, 60(m(3m+1)/10), 60(m(3m+1)/10)+1), m≥ 1,

and (6m-1, 60(m(3m-1)/10), 60(m(3m-1)/10)+1), m≥2.

The question remains as to whether or not there are an infinite number of these two-prime PPT’s.

If in (6m+1, 60(m(3m+1)/10), 60(m(3m+1)/10)+1), m≥ 1, we let m(3m+1)/10=k, k∈ω, we get

(6m+1, 60(m(3m+1)/10), 60(m(3m+1)/10)+1) = (√(1+120k), 60k, 60k+1) and similarly,

if in (6m-1, 60(m(3m-1)/10), 60(m(3m-1)/10)+1), m≥ 2, we let m(3m-1)/10=k, k∈ω, we also get (6m-1, 60(m(3m-1)/10), 60(m(3m-1)/10)+1) = (√(1+120k), 60k, 60k+1).

So, to show that there are an infinite number of such two-prime ppt’s we need to show that

√(1+120k) is an integer and prime, and 60k+1 is also prime, for an infinite number of k∈ω.

If we consider the function defined by f(k) = √(1+120k), mapping [0, ∞) -> [0, ∞), this has a number of obvious integer valued points along it, e.g. f(0)=1, f(1)=11, f(3)=19, f(7)=29, f(8)=31, f(14)=41, f(20)=49, f(29)=59, f(36)=61, f(42)=71, f(52)=79, f(66)=89 …

On closer examination, the f( ) values appear to increase by amounts which follow a simple pattern and that pattern repeats itself indefinitely.

E.g. 11 – 1 =10, 19 – 11 = 8, 29 – 19 = 10, 31 – 29 = 2, 41 – 31 = 10, 49 – 41 = 8, 59 – 49 = 10,

61 – 59 = 2, 71 – 61 = 10, 79 – 71 = 8, 89 – 79 = 10 …

That requires more detailed investigation and proof.

First we will try to define a function g: W -> ω using just the integer valued points from

f(k) = √(1+120k), i.e. such that g(0)=1, g(1)=11, g(2)=19, g(3)=29, g(4)=31, g(5)=41, g(6)=49, g(7)=59 etc.

It is clear from this that we can use g(0)=1, g(1)=11, g(2)=19, g(3)=29 and then for n∈ω (say) define the rest of g( ) recursively using …

g(4n)=g(4n-4)+30

g(4n+1)=g(4n-3)+30

g(4n+2)=g(4n-2)+30

g(4n+3)=g(4n-1)+30

This leads to a simpler definition of g( ) as g(4n)=1+30n, g(4n+1)=11+30n, g(4n+2)=19+30n and g(4n+3)=29+30n, for all n∈ω∪{0}.

Clearly g( ) is an increasing function and so there are an infinite number of values of g( ).

We need to show that these integer g(n) values can be derived from f(k) where k is integer.

Suppose …

(1) f(k)=g(4n)=1+30n. Then √(1+120k)=1+30n or 1+120k=1+60n+900n² . It follows from this that k=n(1+15n)/2. It is easily seen that if n∈ω then so does k and g(k).

(2) f(k)=g(4n+1)=11+30n. The same argument shows that k=1+(11+15n)n/2.

(3) f(k)=g(4n+2)=19+30n. Similarly, k=3+(19+15n)n/2 and,

(4) f(k)=g(4n+3)=29+30n. Hence, k=7+(29+15n)n/2.

It follows therefore that the function f(k) = √(1+120k) has an infinite number of integer points along it, i.e. where k and the corresponding f(k) are both integers.

Note: the integer points along the function f(k) = √(1+120k) can be written in the form

f(n(1+15n)/2) = 1+30n

f((1+3n)(2+5n)/2) = 11+30n

f((2+3n)(3+5n)/2) = 19+30n and

f((1+n)(14+15n)/2) = 29+30n for n∈{0}∪ω.

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Infinite sequences of PPT’s, generated by relatively simple formulae, can be found in a **binary vine-like structure** which sits ‘inside’ the **ternary tree structure** of PPT’s discovered by B. Berggren in 1934.

On reading an excerpt from a book ‘The Pythagorean Theorem: A 4,000-Year History (2007) by Eli Maor’ there is mention of the PPT (4601,4800,6649) which was/is stamped into the clay tablet designated Plimpton 322.

It is straightforward to calculate where this PPT sits within the ‘vine-like’ structure of all PPT’s.

(4601,4800,6649) is the first PPT (n=1) on the branch d=2(c+a) seeded by (704,903,1145) = (a,b,c).

Likewise, (704,903,1145) is the first PPT (n=1) on the branch d=2(c+a) seeded by (21,220,221) = (a,b,c).

(21,220,221) is at position n=10 on the **main stem**.

So, to get to (4601,4800,6649) from the **root (0,1,1)**, take 10 steps along the main stem to (21,220,221), then take one step down the d=2(c+a) branch seeded by (21,220,221) to (704,903,1145), followed by one step down the d=2(c+a) branch seeded by (704,903,1145) to (4601,4800,6649).

**Some identities**

The ‘main stem’ sequence of PPT’s is just (2n+1, n(2n+1)+n, n(2n+1) +n + 1), n∈ω, where 2n+1 is any odd positive integer.

If we choose (a,b,c) = (3,4,5) as a seed PPT, then choose d=2(c+a) we get the generated sequence of PPT’s (16n+4,n(16n+8)-3,n(16n+8)+5) = (16n+4,16n²+8n-3,16n²+8n+5),n∈ω.

So, we have from the main stem, (2n+1)² = (2n² + 2n + 1)² – (2n² + 2n)² , while we also have from (16n+4,16n²+8n-3,16n²+8n+5), n∈ω that (16n²+8n-3)² = (16n²+8n+5)² – (16n+4)².

Now, 16n²+8n-3 = 2(8n²+4n-2) + 1 and so we can replace n by 8n²+4n-2 in

(2n+1)² = (2n² + 2n + 1)² -(2n² + 2n)² to get

(16n²+8n-3)² = (8(4n² +2n -1)² +4(4n² +2n -1) + 1)² -(8(4n² +2n -1)² +4(4n² +2n -1))².

Hence we have …

(16n²+8n+5)² – (16n+4)² = (8(4n²+2n-1)²+4(4n²+2n-1)+1)² – (8(4n²+2n-1)² +4(4n²+2n -1))², n∈ω.

**If n=1 then this becomes 29² – 20² = 221² – 220². **

**If n=10 then we get 1685² – 164² = 1406165² – 1406164².**

Returning to (2n² + 2n + 1)² = (2n+1)² + (2n² + 2n)² and

(16n²+8n+5)² = (16n²+8n-3)² + (16n+4)² if we write 16n²+8n+5 as 2(8n²+4n+2)+1 and substitute

8n²+4n+2 for n in (2n² + 2n + 1)² = (2n+1)² + (2n² + 2n)² we can obtain an identity to generate some Pythagorean quadruples.

This follows …

(8(4n²+2n+1)² + 4(4n²+2n+1)+1)² = (16n+4)² +(16n²+8n-3)² +(8(4n²+2n+1)² + 4(4n²+2n+1))²

**Again if n = 1, we get 421² = 20² + 21² + 420².**

**Also if n=10, 1419613² = 164² + 1677² + 1419612².**

**Fermat’s PPT:** this is the smallest PPT such that the ‘hypotenuse’ and the ‘sum of the other 2 sides’ are squares, i.e. (1061652293520,4565486027761,4687298610289).

A generating sequence linking this back to the main stem is easily derived …

P = 1061652293520

Q = 4565486027761

R = 4687298610289

a = 29729679503

b = 87151633296

c = 92082903025

**n = 4, d=2(c+a)**

P = 29729679503

Q = 87151633296

R = 92082903025

a = 7683073116

b = 10004600587

c = 12614342845

**n = 2, d=2(c-a)**

P = 7683073116

Q = 10004600587

R = 12614342845

a = 142061129

b = 2463588600

c = 2467681129

**n = 1, d=2(c+a)**

P = 142061129

Q = 2463588600

R = 2467681129

a = 1008024

b = 2915143

c = 3084505

**n = 17, d=2(c+a)**

P = 1008024

Q = 2915143

R = 3084505

a = 237943

b = 330576

c = 407305

**n = 2, d=2(c-a)**

P = 237943

Q = 330576

R = 407305

a = 8148

b = 84485

c = 84877

**n = 1, d=2(c-a)**

P = 8148

Q = 84485

R = 84877

a = 75

b = 308

c = 317

**n = 10, d=2(c+a)**

P = 75

Q = 308

R = 317

a = 3

b = 4

c = 5

**n = 4, d=2(c+b)**

Primitive Pythagorean Triples, or PPT’s, are non-trivial integer solutions to the Pythagorean equation z^{2} = x^{2} + y^{2} where we assume that 0<x<y<z, and that x, y and z are co-prime.

Euclid’s formulae allow PPT’s to be generated using 2 (positive integer) parameters, usually denoted by m and n. The PPT’s (usually) take the form ( m^{2}– n^{2}, 2mn, m^{2}+n^{2})

or ( 2mn, m^{2}-n^{2}, m^{2}+n^{2}), where m>n≥1, m and n are co-prime, and m-n is odd. Clearly, m^{2}+n^{2 }is the largest number in each triple.

**Note:** It is easier to spot patterns in generated triples if these are **ordered** as in (x, y, z) where x<y<z.

There are 2 cases that can occur, i.e. ( m^{2}– n^{2}, 2mn, m^{2}+n^{2}) where m^{2}-n^{2}< 2mn and

( 2mn, m^{2}-n^{2}, m^{2}+n^{2}) where 2mn < m^{2}-n^{2}. If m^{2}-n^{2}< 2mn then it is easy to show that

n < m < (1+√2)n. Hence if we wished to write a program to generate such triples we would first have to choose n from 1,2,3, … and then choose m from (n+1), …, int((1+√2)n).

Any proofs that I have seen that establish the above formulae for x, y and z make use of parity, factorisability and co-primeness, but interestingly **do not make use of geometry**.

**It cannot be assumed of course that x, y, and z form the sides of a triangle. That has to be proved.**

However the consequences of proving that x, y, and z form the sides of a triangle lead to a simple method of generating PPT’s which is also simply proved.

**New Method**

**Returning **to the equation z^{2}= x^{2}+ y^{2}, it is easy to show that z<x+y, since x^{2}+ y^{2 }< (x+y)^{2 }and so, since 0<x<y<z, x, y, and z do form the sides of a triangle. It follows, from z<x+y and x<y, that z<2y. Hence y<z<2y, and so there is an integer **a**, where 0<**a**<y, such that z=y+**a**. Also, since 0<x<y, there is an integer **b**, where 0<**b**<y, such that x=y-**b**.

So, z^{2}= x^{2}+ y^{2 }becomes (y+a)^{2} = (y-b)^{2}+ y^{2}, where 0<a<y and 0<b<y.

After multiplying out and re-arranging (y+a)^{2} = (y-b)^{2}+ y^{2 }we get, 2a(a+b) = (y – (a+b))^{2}.

Hence 2|(y-(a+b)) and so y-(a+b) = 2c, where **c** is an integer and it is easy to show that c>0.

**Note:** [2c = y – (a+b) = y – (z-y + y-x) = y + x – z > 0]

So, x = a+2c, y = a + b + 2c = (x+b) and z = 2a + b + 2c = (y+a), where a(a+b)= 2c^{2}.

**Note:** it is easy to show that a<√2 c.

To generate PPT’s, first choose any integral c, **c≥1**; then choose integral **a** in the range **1≤a<√2c **so that a|2c^{2}; then finally choose integral **b** , b≥1, using a(a+b)= 2c^{2},

or b=(2c²)/a – a, making sure that the resulting x, y and z have no common factors.

**For example:**

If c=1 then a=1 and b = 1. Hence we get the PPT (3,4,5).

If c=2 then a=1 and b=7, and this gives the PPT (5,12,13)

If c=3 then a=1 and b=17 **or** a=2 and b=7, giving PPT’s (7,24,25) and (8,15,17).

If c=4 then a=1 and b=31 giving PPT (9,40,41).

If c=5 then a=1 and b=49 **or** a=2 and b= 23 giving PPT’s ( 11, 60, 61) and ( 12, 35, 37).

If c=6 then a = 1 and b = 71 and we get the PPT (13,84,85).

Etc.

The following is a short **Python** program to generate PPT’s using this method.

import math #Generating primitive Pythagorean Triples # def hcf(x,y): #assume y!=x while y!=x: if y>x: y=y-x else: x=x-y #endif #endwhile return x #enddef # def generate_triples(): for c in range(1,60): for a in range(1,int(math.sqrt(2)*c)+1): if int(2*c*c/a)==2*c*c/a: b=int(2*c*c/a)-a x=a+2*c y=a+b+2*c z=2*a+b+2*c if hcf(x,y)==1: print ("c = ",c) print ("x = ",x) print ("y = ",y) print ("z = ",z) print () #endif #endif #endfor #endfor #enddef # generate_triples()

Hence **3d<2a+b** and so **a<3d<2a+b.**

It can also be shown that **4d<2a+b.**

Suppose not, then 4d≥2a+b.

Using the inequality, 4d≥2a+b, in ** we get 3a(a+b)16d≥(6d)^{3} . This simplifies to

2a(a+b)≥9d². From this it follows that (2a+b)² > 18d² or that 16d² > 18d² which is clearly not true.

Hence, **4d<2a+b. ** From this it follows easily that a<(3√2)/2)d = 2.121320344d.

These estimates can be refined a little. For example, if m is the positive root of the equation 36 = m³ + 6m² then **a<md<a+b/2**.

Suppose that md≤a, then from ** we have 3a(a+b)(2a+b)+36a²d<(6d)³ and so

3a(a+b)(2a+b)<36d(6d² – a²). Now md≤a implies that d≤a/m and so,

3a(a+b)(2a+b)<36d(6(a/m)² – a²) = 36a²d(6/m² – 1) = 6a²dm ≤ 6a³. So, 3a(a+b)(2a+b) < 6a³ which is clearly not possible.

Hence** a<md**.

**a+b can’t be a prime number ≥5**. Suppose a+b is a prime p ≥5. Then clearly p|d and so d=pq for some whole number q. So, **a(a+b)(2a+b+12d) = 72d³** becomes

ap(a+p+12pq) = 72p³q³ and reduces to a(a+p+12pq) = 72p²q³. It is clear then that p|a, but this is not possible since a<p. **Hence the result.**

A similar argument shows that **2a+b+12d** can’t be prime.

The equation **a(a+b)(2a+b+12d) = 72d³** can be re-arranged to give …

**a(a+b)(2a+b**) **= 12d(6d² – a(a+b)). **

Clearly 2** |a(a+b)(2a+b) **and so 2 | exactly one of a, (a+b) and (2a+b), otherwise 2 | all three of a, (a+b) and (2a+b) and since 2 | 6d then 2 | x, y and z contradicting gcd(x,y,z)=1. If 2 | a or (a+b) then it is clear that 8 | a or (a+b).

If 2| (2a+b) then **(6d² – a(a+b))** must be odd and so 4| (2a+b) if d is odd and

2^(n+2) | (2a+b) if 2^n | d.

Since a≥1, a+b≥5 and a+b>2d, ** 12d(6d² – a(a+b))=a(a+b)(2a+b) >12d** and so

**(6d² – a(a+b))**>1 or **(6d² – a(a+b))≥2**. Hence **(6d² – a(a+b))** has a prime factor.

If p is prime and p|d then p divides at least one of a, (a+b) and (2a+b). If p| a or (a+b) then p² | a(a+b)(2a+b). If p| any two of a, (a+b) and (2a+b) then it divides all three and since p|d that contradicts gcd(x,y,z)=1. Hence if p|a or (a+b) then p² | a or (a+b) and therefore p³ | a or (a+b). So it follows that d = ABC where a = fA³ , a+b =gB³ and C and a or (a+b) have no common factors. If f or g is 1 then a or (a+b) is a cube. So, suppose that f>1 and so has a prime divisor q (say). Then q is 2, 3 or q|d or q|(6d² – a(a+b)).

The following result is essential to the proof.

If xεR, x≥0, then (1+x/(n-1))^{n-1}≤(1+x/n)^{n}, nεω, n≥2. Equality holds iff x=0. (*)

**Proof**

If xεR, x≥0, and nεω, n≥2, then the following expression may be expanded using the binomial Theorem …

(1+x/n)^{n}-(1+x/(n-1))^{n-1}

= 1 +∑_{1≤r≤n} n(n-1) … (n-r+1)/r! (x/n)^{r}

-(1 +∑_{1≤r≤n-1} (n-1)(n-2) … (n-r)/r! (x/(n-1))^{r})

=∑_{1≤r≤n-1} x^{r}/r![(n-1)/n … (n-r+1)/n – (n-2)/(n-1) … (n-r)/(n-1)]

+ (x/n)^{n}.

It is straightforward to show that …

[(n-1)/n … (n-r+1)/n – (n-2)/(n-1) … (n-r)/(n-1)]>0 since, (n-1)/n > (n-2)/(n-1) … and (n-r+1)/n > (n-r)/(n-1).

Hence it follows that (1+x/n)^{n}-(1+x/(n-1))^{n-1}≥(x/n)^{n} if x≥0.

The result, i.e. (1+x/(n-1))^{n-1}≤(1+x/n)^{n}, follows easily from this, and if equality holds, then (x/n)^{n} = 0, and so x=0.

**Statement** and proof of the Arithmetic-Geometric Mean inequality.

For each nεω, let P(n) be the proposition that for any set of non-negative numbers a_{1}, a_{2}, … a_{n},

(a_{1} + a_{2} + …. + a_{n})/n ≥(a_{1}a_{2} … a_{n})^{1/n}

with equality if and only if a_{1} = a_{2} = a_{3} … = a_{n}.

**Proof**

P(1) is trivially true, i.e. a_{1}/1 = a_{1}^{1/1}.

P(2) If a_{1} and a_{2} are ≥0 then it is easily shown that …

(a_{1} + a_{2})^{2}≥ 4a_{1}a_{2} and so (a_{1} + a_{2})/2 ≥ √(a_{1}a_{2}).

If (a_{1}+ a_{2})/2 = √(a_{1}a_{2}) then (a_{1} – a_{2})^{2} = 0 and so a_{1} = a_{2}.

Now suppose that P(n-1) is true for n-1 ε ω and n – 1≥ 2.

Suppose that a_{1}, a_{2}, … a_{n} are n non-negative numbers. We can assume that a_{1}≤a_{2}≤a_{3} …≤ a_{n}. If not then we can choose a re-arrangement of a_{1}, a_{2}, … a_{n} that does have this property. It now follows that …

(a_{2} + a_{3} + … a_{n})/(n-1)≥(a_{2}a_{3} … a_{n})^{1/(n-1)}or

a_{2}a_{3}… a_{n}≤((a_{2}+ a_{3}+ … a_{n})/(n-1))^{(n-1)} with equality iff a_{2} = a_{3} = a_{4} … = a_{n}.

Now let D = (a_{2} – a_{1}) + (a_{3} – a_{1}) + … + (a_{n} – a_{1}). Clearly D≥0 with equality iff a_{1}=a_{2}=a_{3} … = a_{n}.

D = a_{2} + a_{3} + … +a_{n} – (n-1)a_{1} and so, a_{1} + D/(n-1) = (a_{2} + a_{3} + … + a_{n})/(n-1).

So, a_{2}a_{3}… a_{n}≤(a_{1} + D/(n-1))^{(n-1)} with equality iff a_{2}= a_{3}= a_{4}… = a_{n}.

Hence, multiplying by a_{1} we get a_{1}a_{2}a_{3}… a_{n}≤a_{1}(a_{1}+ D/(n-1))^{(n-1)}.

Applying (*) above we get a_{1}a_{2}a_{3}… a_{n}≤a_{1}(a_{1}+ D/(n-1))^{(n-1)}≤(a_{1}+ D/n)^{n}, or

a_{1}a_{2}a_{3}… a_{n}≤(a_{1}+ D/n)^{n}=((a_{1} + a_{2} + … a_{n})/n)^{n}, with equality iff D=0 or iff a_{1}=a_{2}=a_{3}… = a_{n}.

Hence it follows that (a_{1} + a_{2} + … +a_{n})/n≥ ^{n}√(a_{1}a_{2} … a_{n}) with equality iff a_{1}=a_{2}=a_{3}… = a_{n}. Thus P(n) is true and the result follows by the principle of Mathematical Induction.

**Applications of the AGM to the sequence n ^{1/n}, where nεω, n≥1.**

Numerical computation of the first few terms of this sequence shows that the first term 1^{1/1} = 1, the second is 2^{1/2} = √2 and the third term is ^{3}√3 ( the maximum value of the sequence). For n≥3 the sequence appears to be monotone decreasing, tending to a limit of 1.

Firstly, 2 subsequent values cannot be equal. Otherwise n^{1/n} = (n+1)^{1/(n+1)} for some n. From this equation (if true) it follows that n^{(n+1)} = (n+1)^{n}. If n=1 then it follows that 1 = 2 which is obviously false. If n>1 then n and n+1 must have a common prime factor, again false since n and n+1 are coprime.

So, for all nεω, n≥1, either n^{1/n}> (n+1)^{1/(n+1) }or n^{1/n}< (n+1)^{1/(n+1)}.

Suppose that n^{1/n}> (n+1)^{1/(n+1)}. Then, n^{(n+1)}> (n+1)^{n}.

So, n^{(n+1)}– n^{n}> (n+1)^{n}-n^{n }and from this it follows that …

n^{n}(n-1)> (n+1)^{n-1} + (n+1)^{n-2}n + … + n^{n-1} >n(n(n+1))^{(n-1)/2} (applying the AGM).

Simplifying n^{n}(n-1)>n(n(n+1))^{(n-1)/2}gives n-1>((n+1)/n)^{(n-1)/2}>1. Hence it follows that n≥3.

**So, for all nεω, n≥3, n ^{1/n}> (n+1)^{1/(n+1)}.**

To prove that lim _{n→∞ }n^{1/n }= 1, first observe that n^{1/n}> 1 for all n> 1, otherwise n^{1/n}≤1 for some n>1. n^{1/n}≤ 1 implies that n≤1, which clearly gives a contradiction, so n^{1/n}> 1 for all n> 1.

Next, if n>2, we apply the AGM inequality to the set of n positive numbers 1, 1, … ,1 (n-2 times),√n,√n.

This gives (n-2+2√n)/n>n^{1/n}, or 1-2/n+2/√n>n^{1/n}. This last inequality implies that as n→∞, n^{1/n} approaches arbitrarily close to 1. Hence the result **lim _{n→∞ }n^{1/n}= 1**.

By the Fermat equation I mean, of course, **z ^{n} = x^{n} + y^{n }** where x, y and z are positive integers and n>=2 is also a positive integer. The solutions to the equation in the case n=2 are well known and can be generated by the application of

When n=2 the solutions to z^{n}= x^{n}+ y^{n }are called Pythagorean Triples and, if these have no common factors, they are called **Primitive Pythagorean Triples**, e.g. (3,4,5).

If we are going to look for solutions to these equations (or show there are none) then we can make some simplifying assumptions without loss of generality, e.g. that x, y and z have no common factors and that 0<x<y<z.

From z^{n}= x^{n}+ y^{n }we have that z^{n} – y^{n} = x^{n} and so z-y = x^{n}/(z^{(n-1)} + z^{(n-2)}y+ …. y^{(n-1)}).

The denominator of the RHS of this last equation, i.e. z^{(n-1)}+ z^{(n-2)}y+ …. y^{(n-1)}, can easily be shown to be greater than ny^{(n-1) }(since z > y) and so we have that …

z-y<x^{n}/(ny^{(n-1)}) <x/n (since (x/y)^{(n-1)}<1). Hence x>n(z-y), and so x≥n+1

(since z>y and x, y, z and n are positive integers). This in turn means that y≥n+2 and z≥n+3.

Various other estimates(inequalities) can be derived involving x, y and z e.g. **z<x ^{2}**.

**Proof** (that **z<x ^{2}**).

If we assume that z≥x^{2} then the case z=x^{2} contradicts the assumption that x, y and z have no common factors. So that can be dismissed.

If z>x^{2} then zx^{(n-2)}>x^{n}=z^{n}-y^{n}.

Hence z-y<zx^{(n-2)}/(z^{(n-1)} + … + zy^{(n-2)}+y^{(n-1)}) <1. So it must be the case that z≤y which contradicts z>y.

**It follows therefore that z<x ^{2}.**

(Alternative proof)

It is straight-forward to prove that z^{n-1}<x^{n}, and from that result z<x^{2} follows.

If z^{n-1}<x^{n} is not true then z^{n-1}≥x^{n }and so x^{n}=z^{n}-y^{n}≤z^{n-1}.

From this it follows that z-y≤z^{n-1}/(z^{(n-1)}+ … + zy^{(n-2)}+y^{(n-1)})<1 and so z≤y which contradicts y<z. **Hence, z ^{n-1}<x^{n }and so z<(x/z)^{n-2}x^{2}<x^{2}**.

y an z can now be written as z=ax+b and y=cx+d where 0<a, b, c, d<x (as if we are working in base x). So we have (ax+b)^{n}=x^{n}+(cx+d)^{n} .

The primitive Pythagorean triples where z≤100 are ( 3 , 4 , 5 ) ( 5, 12, 13 ) ( 7, 24, 25)

( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29)

(28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).

If we pick out the first few triples where x is odd we get (3 , 4 , 5 ) (5, 12, 13 ) (7, 24, 25)

(9, 40, 41) (11, 60, 61) etc. and if the position in this sequence is denoted by **n**,

the general term is given by (2n+1, n(2n+1)+n, n(2n+1)+(n+1)) where n∈ω.

Similarly if we pick out (8,15,17) (16,63,65) (24,143,145) etc. and again denote then general position in this sequence by **n** we get (8n, (2n-1)8n+8n-1, (2n)8n+1) where n∈ω.

Likewise with ((16n-4), (4n-2)(16n-4)+(16n-5),(4n-1)(16n-4)+1) which generates the sequence that starts off (12,35,37), (28,195,197), (44,483,485) etc. and

((16n+4),4n(16n+4)+16n+3,(4n+1)(16n+4)+1) which generates

(20,99,101), (36, 323,325), (52,165,173) etc.

**Note:** ((16n+4),n(16n+4)+(4n-3),n(16n+4)+(4n+5)) where n∈ω also generates Pythagorean triples such as

(20,21,29), (36,77,85), (52,165,173) etc. and because

((16n+4),4n(16n+4)+16n+3,(4n+1)(16n+4)+1) has the same ‘x’ value for each n, we can state that …

(16n+4)^{2}=((4n+1)(16n+4)+1)^{2} – (4n(16n+4)+(16n+3))^{2}

= (n(16n+4)+(4n+5))^{2}– (n(16n+4)+(4n-3))^{2} for all n∈ω.

Then, if for example if n=4 …

68^{2}= 1157^{2}-1155^{2}= 293^{2}-285^{2} and so 1157^{2}+285^{2}= 1155^{2}+293^{2}.

These sequences don’t account for all primitive Pythagorean triples, and no doubt there are others, but clearly they all fit the pattern (x,cx+d,ax+b) .

The question remains as to what can be proved about a, b, c and d in the cases where n>=3 or (indeed n=2).

**The inequality, z<x ^{2}, can be sharpened a little to z<x^{2} – (n-1)x.**

**Proof**

Suppose that z>x^{2}– (n-1)x. It follows that x^{2}<z+(n-1)x and so x^{n}<(z+(n-1)x)x^{n-2} .

Hence, z^{n}-y^{n}<(z+(n-1)x)x^{n-2} and so …

z-y<((n-2)x^{n-1}+zx^{n-2}+x^{n-1})/(z^{n-1}+ …+zy^{n-2}+y^{n-1})<1.

From z-y<1 it follows again that z≤y, which contradicts one of the initial assumptions that y<z. Hence the inequality z<x^{2}– (n-1)x is proved.

**Note:** the case z=x^{2}– (n-1)x cannot occur because x and z would then have a common factor, again contradicting one of the initial assumptions.

After giving this last result some thought, I wondered if in fact z<x^{2}– (n-1)y?

This is also true, but only if n≥3, and can be proved similarly to the above.

What about z<x^{2}– (n-1)z or z<x^{2}/n?

This is also true, but again only if n≥3.

**Proof**

Suppose that z<x^{2}/n is false, i.e. that z≥x^{2}/n. Then x^{2}≤nz. Hence, x^{n}≤nzx^{n-2} and so z^{n}-y^{n}≤nzx^{n-2}.

It follows then that z-y≤nzx^{n-2}/(z^{n-1} + yz^{n-2} + … y^{n-1}). We need to show now that the RHS of this inequality is <1, or that the denominator of the RHS is >nzx^{n-2}.

To this end suppose then that, for n≥3, z^{n-1}+ yz^{n-2}+ … y^{n-1}≤nzx^{n-2}.

By applying the AGM inequality to nzx^{n-2}≥z^{n-1}+ yz^{n-2}+ … y^{n-1} we get …

nzx^{n-2}>n(zy)^{(n-1)/2}. This reduces to zx^{n-2}>(zy)^{(n-1)/2}. Squaring both sides gives z^{2}x^{2n-4}>(zy)^{n-1} and so x^{2n-4}> z^{n-3}y^{n-1}. This reduces to z^{n-3}<x^{n-3}(x/y)^{n-1}<x^{n-3 }which is clearly false for n≥3.

Hence, for n≥3, z^{n-1}+ yz^{n-2}+ … y^{n-1}>nzx^{n-2} and so z-y<1 or z≤y which contradicts one of the initial assumptions. So z<x^{2}/n, if n≥3, is proved.

**Note:** when n=2, z<x^{2}/2 is true **for some triples** such as (8,15,17) but **false for others** such as (3,4,5).

It has already been shown that x≥n+1, y≥n+2 and z≥n+3. z<x^{2}/n means then that x^{2}≥n(n+3)+1 or that **x≥n+2 **if n≥3**.**

**Euclid’s Formulae and Primitive Pythagorean Triples (PPT’s).**

These formulae allow PPT’s to be generated using 2 (positive integer) parameters, usually denoted by m and n. The PPT’s (usually) take the form ( m^{2} – n^{2}, 2mn, m^{2}+n^{2}) where m>n≥1, m and n are coprime and m-n is odd. Clearly, m^{2}+n^{2} is the largest number in each triple. It is easier to spot patterns in generated triples if these are ordered as in (x, y, z) where x<y<z.

Hence there are 2 cases that can occur, i.e. (m^{2}– n^{2}, 2mn, m^{2}+n^{2}) where m^{2}-n^{2} < 2mn and

(2mn, m^{2}-n^{2},m^{2}+n^{2}) where 2mn < m^{2} -n^{2}.

Notation: Let the set of all PPT’s (x,y,z) where x<y<z be denoted by PPT and let the set of PPT’s where x is odd be denoted by OEO while when x is even the set of PPT’s is denoted by EOO.

Clearly PPT = OEO ∪ EOO and OEO ∩ EOO =Ø.

If m^{2}-n^{2}< 2mn then it is easy to show that n < m < (1+√2)n. Hence if we wished to write a program to generate such triples we would first have to choose n from 1,2,3, … and then choose m from (n+1), …, int((1+√2)n). I have written such a program in Python, where z and y, following integer division by x, are printed in the form ax + b and cx + d respectively.

Here is the program followed by its output …

#Primitive Pythagorean Triples in the case m*m-n*n<2mn import math # def gcd(a,b): while a!=b: if a>b: a=a-b else: b=b-a #endif #endwhile return a #enddef # for n in range(1,7): lower=n upper=int((1+math.sqrt(2))*n) for m in range(lower+1,upper+1): if gcd(m,n)==1 and gcd(m-n,2)==1: # x=m*m-n*n y=2*m*n # c=int(y/x) d=y-c*x # z=m*m+n*n a=int(z/x) b=z-a*x # first_form="("+str(x)+","+str(y)+","+str(z)+")" second_form="("+str(x)+","+"["+str(c)+"*"+str(x)+"+"+str(d)+"]"+","+"["+str(a)+"*"+str(x)+"+"+str(b)+"]"+")" print (first_form.ljust(20," ")) print (second_form.ljust(40," ")) print () #endif #endif #endfor

(3,4,5)

(3,[1*3+1],[1*3+2])

(5,12,13)

(5,[2*5+2],[2*5+3])

(7,24,25)

(7,[3*7+3],[3*7+4])

(9,40,41)

(9,[4*9+4],[4*9+5])

(33,56,65)

(33,[1*33+23],[1*33+32])

(65,72,97)

(65,[1*65+7],[1*65+32])

(11,60,61)

(11,[5*11+5],[5*11+6])

(39,80,89)

(39,[2*39+2],[2*39+11])

(119,120,169)

(119,[1*119+1],[1*119+50])

(13,84,85)

(13,[6*13+6],[6*13+7])

(85,132,157)

(85,[1*85+47],[1*85+72])

(133,156,205)

(133,[1*133+23],[1*133+72])

Using the first 4 triples it can be easily seen that these satisfy the general term (2n+1,n(2n+1)+n,n(2n+1)+(n+1)).

However, the triple (33,[1*33+23],[1*33+32]) is just the first of an infinite sequence of triples that satisfy (18n+15,n(18n+15)+(15n+8),n(18n+15)+(15n+17)) where n∈ω.

Such sequences are not at all obvious if PPT’s are generated using the usual form of Euclid’s formulae and if y and z are not written as cx+d and ax+b respectively.

There are other patterns that emerge that could also merit some analysis.

If m^{2}-n^{2 }>2mn then it is easy to show that m > (1+√2)n. Hence if we wished to write a program to generate such triples we would first have to choose n from 1,2,3, … and then choose m from int((1+√2)n)+1, int((1+√2)n)+2,…

Here is such a program followed by its output in the same format as the above …

#Primitive Pythagorean Triples in the case m*m-n*n>2mn import math # def gcd(a,b): while a!=b: if a>b: a=a-b else: b=b-a #endif #endwhile return a #enddef # for n in range(1,5): bandwidth=10 lower=int((1+math.sqrt(2))*n) for m in range(lower+1,lower+bandwidth): if gcd(m,n)==1 and gcd(m-n,2)==1: # y=m*m-n*n x=2*m*n # c=int(y/x) d=y-c*x # z=m*m+n*n a=int(z/x) b=z-a*x # first_form="("+str(x)+","+str(y)+","+str(z)+")" second_form="("+str(x)+","+"["+str(c)+"*"+str(x)+"+"+str(d)+"]"+","+"["+str(a)+"*"+str(x)+"+"+str(b)+"]"+")" print (first_form.ljust(20," ")) print (second_form.ljust(40," ")) print () #endif #endif #endfor

(8,15,17)

(8,[1*8+7],[2*8+1])

(12,35,37)

(12,[2*12+11],[3*12+1])

(16,63,65)

(16,[3*16+15],[4*16+1])

(20,99,101)

(20,[4*20+19],[5*20+1])

(20,21,29)

(20,[1*20+1],[1*20+9])

(28,45,53)

(28,[1*28+17],[1*28+25])

(36,77,85)

(36,[2*36+5],[2*36+13])

(44,117,125)

(44,[2*44+29],[2*44+37])

(52,165,173)

(52,[3*52+9],[3*52+17])

(48,55,73)

(48,[1*48+7],[1*48+25])

(60,91,109)

(60,[1*60+31],[1*60+49])

(84,187,205)

(84,[2*84+19],[2*84+37])

(96,247,265)

(96,[2*96+55],[2*96+73])

(88,105,137)

(88,[1*88+17],[1*88+49])

(104,153,185)

(104,[1*104+49],[1*104+81])

(120,209,241)

(120,[1*120+89],[2*120+1])

(136,273,305)

(136,[2*136+1],[2*136+33])

It is easy to show that these satisfy (4n+4,n(4n+4)+(4n+3),(n+1)(4n+4)+1) where n∈ω.

Further on in the output we see …

(20,[1*20+1],[1*20+9])

(28,[1*28+17],[1*28+25])

(36,[2*36+5],[2*36+13])

(44,[2*44+29],[2*44+37])

(52,[3*52+9],[3*52+17])

(60,[3*60+41],[3*60+49])

(68,[4*68+13],[4*68+21])

(76,[4*76+53],[4*76+61])

(84,[5*84+17],[5*84+25])

(92,[5*92+65],[5*92+73])

(100,[6*100+21],[6*100+29])

(108,[6*108+77],[6*108+85])

From these we get the sequences (16n+4,n(16n+4)+(4n-3),n(16n+4)+4n+5), where n∈ω,

and (16n+12,n(16n+12)+12n+5),n(16n+12)+(12n+13)) where n∈ω and from …

(48,[1*48+7],[1*48+25])

(60,[1*60+31],[1*60+49])

(84,[2*84+19],[2*84+37])

(96,[2*96+55],[2*96+73])

(120,[3*120+31],[3*120+49])

(132,[3*132+79],[3*132+97])

(156,[4*156+43],[4*156+61])

(168,[4*168+103],[4*168+121])

(192,[5*192+55],[5*192+73])

(204,[5*204+127],[5*204+145])

we get (36n+12,n(36n+12)+(12n-5),n(36n+12)+(12n+13)) where n∈ω and

(36n+24,n(36n+24)+(24n+7),n(36n+24)+(24n+25)) where n∈ω.

On the face of it there seem to be many more examples of infinite sequences of PPT’s from the case

m^{2}– n^{2}>2mn than from m^{2}– n^{2 }<2mn. However, looking at the above examples, there does seem to be a crossover between the two cases, e.g. in the appearance of (7,24,25) in

(36n+24,n(36n+24)+(24n+7),n(36n+24)+(24n+25)) etc.

In fact it is straightforward to show that if (a,b,c) is a primitive Pythagorean triple from the set OEO and if d=2(c-a) it follows that …

**(dn+b,n(dn+b)+(bn+a),n(dn+b)+(bn+c)),** where n∈ω, will generate an infinite sequence of PPT’s in the set EOO.

**Proof**

It is easy to show that the 3 values in (dn+b,n(dn+b)+(bn+a),n(dn+b)+(bn+c)), where n∈ω, are in ascending order and that dn+b is **even** while n(dn+b)+(bn+a) and n(dn+b)+(bn+c) are **odd** where n∈ω. Also …

(n(dn+b)+(bn+c))^{2} – (n(dn+b)+(bn+a))^{2} = (c-a)(2n(dn+b)+2bn+c+a)

=dn(dn+b)+dbn+c^{2}-a^{2} = dn(dn+b)+dbn+b^{2} = (dn+b)^{2}. (Using c^{2}=a^{2}+b^{2})

It is also true that if (a,b,c) is a primitive Pythagorean triple from the set OEO, and if d=2(c+a), then, as in the above proof, (dn+b,n(dn+b)+(bn-a),n(dn+b)+(bn+c)), n∈ω will generate an infinite sequence of PPT’s in the set where EOO.

So, every (a,b,c) in the set OEO can be used to generate **2 infinite sequences** of PPT’s where the generated values lie in the set where EOO.

It is also clear that for every (a,b,c) in the set EOO, 2 infinite sequences of PPT’s can be generated using exactly the same formulae, i.e.

(dn+b,n(dn+b)+(bn+a),n(dn+b)+(bn+c)) where n∈ω, d=2(c-a) and

(dn+b,n(dn+b)+(bn-a),n(dn+b)+(bn+c)) where n∈ω, d=2(c+a),

and all of the generated PPT’s lie in the set where OEO.

*****************************************************************************************

These infinite sequences of PPT’s can be arranged in a **binary vine-like structure** rather than a ternary tree. The ternary tree structure of PPT’s was discovered by B. Berggren in 1934. However, that hierarchical structure contains this binary vine-like structure within it.

The (horizontal) **main stem** of this vine-like structure has its **root** in the **degenerate triple (0,1,1)**. This gives rise to the generating formulae (2n+1, n(2n+1)+n, n(2n+1)+(n+1)) where n∈ω and therefore …

(0,1,1) —> (3,4,5) —> (5,12,13) —> (7,24,25) —> (9,40,41) —> (11,60,61) —> (13,84,85) —> (15,112,113) —> (17,144,145) —> (19,180,181) —> (21,220,221) —> etc.

**Note:** each one of these triples belongs to the set OEO. Also, if (a,b,c)=(0,1,1) it does not matter if d=2(c+a) or d=2(c-a) the resulting sequence is the same.

Each of these PPT’s will then generate 2 infinite sequences in the set EOO using the formulae …

(dn+b,n(dn+b)+(bn+a),n(dn+b)+(bn+c)) where n∈ω, where d=2(c-a) and …

(dn+b,n(dn+b)+(bn-a),n(dn+b)+(bn+c)) where n∈ω, where d=2(c+a).

**For example** from (3,4,5) we get 2 infinite sequences which start off …

(8,15,17) —> (12,35,37) —> (16,63,65) —> (20,99,101) —> (24,143,145) —> (28,195,197) —> etc and …

(20,21,29) —> (36,77,85) —> (52,165,173) —> (68,285,293) —> (84,437,445) —> (100,621,629) —> etc.

**These could be arranged to hang down from (3,4,5) but also map across into the set EOO.**

Repeating the process for each triple in these sequences gives 2 further sequences belonging to OEO.

So, (8,15,17) generates (33,56,65) —> (51,140,149) —> (69,260,269) —> etc. and …

(65,72,97) —> (115,252,277) —> (165,532,557) —> etc.

**Further sequences of PPT’s can be generated using the above formulae by interchanging a and b.**

If (a,b,c) is a PPT where a<b<c, (dn+a,n(dn+a)+(an+b),n(dn+a)+(an+c)) where n∈ω and d=2(c-b) generates another infinite sequence of PPT’s. The difference here is that **if (a,b,c) ∈ EOO** then

**so do all the members of (dn+a,n(dn+a)+(an+b),n(dn+a)+(an+c)), n∈ω.** Similarly, if (a,b,c) ∈ OEO then so do all the members of (dn+a,n(dn+a)+(an+b),n(dn+a)+(an+c)), n∈ω.

More sequences can be generated by (dn+a,n(dn+a)+(an-b),n(dn+a)+(an+c)), n∈ω and d=2(c+b), **but with a slight difference.**

We need to ensure that dn+a<n(dn+a)+(an-b)<n(dn+a)+(an+c)). Clearly the second inequality is true for all n∈ω.** However, the first inequality is not necessarily true for all n∈ω.**

We need to ensure that dn+a<n(dn+a)+(an-b) or 0<dn² + (2a-d)n – (a+b).

Solving this inequality leads to n>(1+√2)/2 – (c-b)/(2a). From c² = a² + b² and a<b<c, it follows easily that c>√2 a. Also, c – b = a² / (c+b) < a/(√2 + 1) = a(√2 – 1).

**Hence it follows that n>(1+√2)/2 – (√2 – 1)/2 =1.**

So, if (a, b, c) is a PPT where a<b<c, and d=2(c+b), (dn+a,n(dn+a)+(an-b),n(dn+a)+(an+c)) will generate an infinite sequence of PPT’s,** if n∈ω and n>1.**

As in the case where d=2(c-b), if (a,b,c)∈ EOO and d=2(c+b), (dn+a,n(dn+a)+(an-b),n(dn+a)+(an+c)) belongs to EOO if n∈ω and n>1, and if (a,b,c)∈ OEO, so also do (dn+a,n(dn+a)+(an-b),n(dn+a)+(an+c)) againif n∈ω and n>1.

It remains to be proved that every PPT (P,Q,R) (say) either belongs to the **horizontal main stem** or that there is a ‘smaller’ PPT (a,b,c) which generates an infinite sequence of PPT’s containing (P,Q,R).

**Proof**

Suppose that (P,Q,R) is **any** PPT where 0<P<Q<R, R² = P² + Q² and gcd(P,Q,R)=1.

Define n = int(Q/P), d=2(R – Q), b = P – nd, c = R – n(P+b) and a=abs(Q-n(P+b)).

Clearly, n, a, b, c and d are integers and n≥1, a≥0 and d≥2.

**Note:** n either represents the position along the main stem if (P,Q,R) belongs to the main stem, i.e. if (P,Q,R) has been generated from the degenerate PPT (0,1,1), or n represents the depth of (P,Q,R) down one of the** four possible sequences** generated by a ‘smaller’ non-degenerate PPT (a,b,c).

Now, n = int(Q/P) implies that n<Q/P<n+1, or nP<Q<nP+P, otherwise the initial assumption that gcd(P,Q,R)=1 is contradicted.

Also, d=2(R – Q) implies that d/2 = R-Q = P²/(R+Q) < P²/(2Q). Hence, P²>dQ and so P>d(Q/P)>nd.

So it follows that b = P – nd>0 and therefore **b>0**.

Next, from d/2 =P²/(R+Q) we get d/2 > P²/(2R) or R>P²/d.

So, R>(P/d)(nd+b)=nP +Pb/d>nP+nb=n(P+b). Hence, c=R-n(P+b)>0 or **c>0**.

Now, **c² – b²** = (R – n(P + b))² – (P – nd)² = R² – 2Rn(P+b) + n²(P+b)² – P² + 2Pnd – n²d²

= Q² + n²(P+b)² – (2Q + d)n(P+b) + 2Pnd – n²d²

= Q² + n²(P+b)² – 2Qn(P+b) – nd(P+b)+ 2Pnd – n²d²

=(Q – n(P+b))² -nd(P+b -2P +nd)

= (Q – n(P+b))²

= **a²**.

Hence it follows that **c² = a² + b²**.

We now look at the cases (1) a = 0 and (2) a≠ 0 (basically a>0).

(1) If a = 0 then c = b or c = – b. a = 0 also implies that Q=n(P+b).

If c = – b then c<0 which is clearly not true.

So it follows therefore that c = b. Now b = c = R – n(P+b) = R – Q = d/2 (since a=0).

Then P = b+nd = b(1+2n), Q= n(P+b) = 2bn(1+n), and so R = b(2n(n+1)+1). Clearly then b=1 since b is a common divisor of P, Q and R and gcd(1+2n, 2n(1+n), 2n(n+1)+1) = 1.

**Hence a = 0 implies that c = b = 1 and so (P,Q,R) belongs to the main stem sequence.**

(2) a≠0. Clearly a, b and c form a PPT otherwise they have a common factor and so gcd(P,Q,R)≠1.

We need to look at 2 cases, Q-n(P+b)>0 and Q-n(P+b)<0.

(2.1) **Suppose a = Q-n(P+b)>0.** Then d/2=R-Q=c-a<b since c<a+b. So, **d/2<b**.

Conversely, if d/2<b then R-Q=d/2<b. Hence, n(P+b) +c – Q<b and so 0<c-b<Q-n(P+b).

(2.2) **Suppose that a = -Q+n(P+b).** Then d/2=R-Q=c+a>c>b. So, **d/2>b**.

Conversely if d/2>b then d/2=R-Q=c+a>b and so (c-a)(c+a)>(c-a)b and using c² – a² = b²

we have that b²>(c-a)b which implies that b>c-a and so a>c-b.

Hence it follows that -Q+n(P+b)>0.

So, once d and b have been determined from P, Q and R the sign (+ or -) of d/2-b determines the correct expression to use for a.

We need to show now that, in both cases where a>0, that a<b.

Clearly a≠b (a>0) otherwise c would be irrational, so either a<b or b<a.

We need to prove that a<b either by a contradiction argument or perhaps find a direct proof that a<b.

**Proof**

[1] We start with the case where a>0 and d/2<b. This means that Q=n(P+b)+a and that d/2=c-a.

We will assume that **b<a**.

From the definition of n, i.e. that n=int(Q/P) it follows that Q<(n+1)P.

So, n(P+b)+a=Q<(n+1)P and so nb+a<P. Using b<a now we have nb+b<P=nd+b, or b<d=2(c-a).

b<2(c-a) implies that b(c+a)<2b² or that c+a<2b from which it follows that a<b. This contradicts the initial assumption that b<a. Hence it follows that **a<b**.

[2] We start again with the second case where a>0 but d/2>b. This means that Q=n(P+b)-a and that d/2=c+a.

Now n≥1. If n=1 then Q=P+b-a>P and so b>a.

If n>1 then (P-d,Q-(2n-1)d-2b,R-(2n-1)d-2b) = ((n-1)d+b,(n-1)((n-1)d+2b)-a,(n-1)((n-1)d+2b)+c) is also a PPT where P-d<Q-(2n-1)d-2b<R-(2n-1)d-2b). This follows from d/2>b.

(P-d,Q-(2n-1)d-2b,R-(2n-1)d-2b) has the same form as (P,Q,R) except that n has been replaced by (n-1). If n=2 then it follows that b>a as above.

If we repeat this process, eventually we must end up with either b>a or the PPT (d+b,d+2b-a,d+2b+c) where

d+b<d+2b-a<d+2b+c, i.e. b>a. Hence the result is proved.

**Note:** It is clear that if a>0 then c<R and so (a,b,c) is a ‘smaller’ PPT that (P,Q,R).

The result is also true if a and b are interchanged in the generating formulae.

**Hence the overall result is proved**, i.e. that every PPT (Primitive Pythagorean Triple) (P,Q,R) belongs to either the ‘main stem’ sequence (1+2n, 2n(1+n), 2n(n+1)+1) for some positive integer n, or, there is a smaller PPT (a,b,c) such that (P,Q,R) belongs to at least one of the **four **possible sequences defined by …

(dn+b, dn²+2bn+a,dn²+2bn+c), n∈ω where d=2(c-a),

(dn+b, dn²+2bn-a,dn²+2bn+c),n∈ω where d=2(c+a),

(dn+a,n(dn+a)+(an+b),n(dn+a)+(an+c)) where n∈ω and d=2(c-b) or

(dn+a,n(dn+a)+(an-b),n(dn+a)+(an+c)) where n∈ω, n>1, and d=2(c+b).

**Note:** Some PPT’s belong to 2 different sequences, e.g. (1365,5428,5597) can be generated from (13,84,85) using d=2(c+b) and n=4 in (dn+a,n(dn+a)+(an-b),n(dn+a)+(an+c)) while …

(1365,5428,5597) can also be generated from (280,351,449) using d=2(c-a), n=3 and

(dn+b, dn²+2bn+a,dn²+2bn+c).

***********************************************************************************

**Note:** A ‘Silver Ratio’ PPT (a,b,c) is one where b-a=1. From the PPT sequence (dn+b,n(dn+b)+(bn-a),n(dn+b)+(bn+c)), where d=2(c+a), if we set n=1 we can generate another ‘Silver Ratio’ PPT which can be simplified to (2a+b+2c,a+2b+2c, 2a+2b+3c). Clearly, (a+2b+2c) – (2a+b+2c) = b-a. Hence there are an **infinite number** of these ‘Silver Ratio’ PPT’s .

**Further inequalities** flowing from z^{n} = x^{n} + y^{n} include …

x^{3}>ny^{2}, if n≥3 and the stronger x^{3}>nyz, if n≥3.

**Proof that x ^{3}>nyz, if n≥3.**

Suppose not then, x^{3}<nyz, if n≥3.

So, it follows that x^{n}<nyzx^{(n-3)} and so z-y<nyzx^{(n-3)}/(z^{n-1}+ yz^{n-2}+ … +y^{n-1})<nyzx^{(n-3)}/(n(yz)^{(n-1)/2})<(x^{2}/(yz))^{(n-3)/2}≤1.

Hence z-y<1 which is not possible, and so the result is proved.

**Note:** From x^{3}>nyz and using z≥y+1 and y≥x+1 we can derive x^{3}>ny^{2}+nx+n, which is clearly an inequality based on the form of an elliptic curve.

**What about x ^{3}>nz^{2}, if n≥3???**

If we try to prove x^{3}>nz^{2} where n≥3 using proof by contradiction, then we would start by assuming that x^{3}<nz^{2}. It would follow then that x^{n}<nz^{2}x^{n-3}.

By applying the AGM inequality as before we get z-y<nz^{2}x^{n-3}/(n(yz)^{(n-1)/2}).

This can be simplified to z-y<(z/y)(x^{2}/yz)^{(n-3)/2} ≤ z/y.

Hence z-y<z/y and so z(y-1)≤y^{2} – 1 or z≤y + 1 (since y>1).

This in turn means that z=y+1 and so the Fermat equation becomes

(y+1)^{n} = y^{n} + x^{n}. Consequences ??? …

**Further inequalities** easily proved from the Fermat equation include …

n(yz)^{(n-1)/2} < x^{n}, from which x^{3}>nyz can be deduced, and 2n(xz)^{(n-1)/2} <y^{n}.

From these it follows easily that 2n^{2}z^{(n-1)} < (xy)^{(n+1)/2} and hence 2n^{2}z^{(n-1)}<y^{(n+1)} and so z>√2n. Using z>√2n in 2n^{2}z^{(n-1)}<y^{(n+1)} also means that y>√2n.

*****************************************************************************************

**Returning** to the equation z^{n}= x^{n}+ y^{n}, n≥2, it is easy to show that z<x+y<2y under the assumptions made at the start of this article. Hence, z=y+a, where 0<a<y. Also, x=y-b, where 0<b<y.

So, z^{n}= x^{n}+ y^{n} becomes (y+a)^{n} = (y-b)^{n} + y^{n}, where 0<a<y and 0<b<y.

When n=2, and after multiplying out and re-arranging we get, 2a(a+b) = (y -(a+b))^{2}.

Hence 2|(y-(a+b)) and so y-(a+b) = 2c, where it is easy to show that c>0.

So, x = a+2c, y = a + b + 2c = (x+b) and z = 2a + b + 2c = (y+a), where a(a+b)= 2c^{2}.

To generate PPT’s, first choose integral **c>0**. Then choose **a** so that a|2c^{2}; then finally choose b using a(a+b)= 2c^{2 }making sure that x, y and z have no common factors.

If c=1 then a=1 and b = 1. Hence we get the PPT (3,4,5).

If c = 5 then a(a+b) = 50, and so we can have a = 1, b = 49 **or** a = 2, b = 23.

These give the PPT’s (11, 60, 61) and (12, 35, 37) respectively.

Similarly, if c = 6, then from a = 1 and b = 71 we get the PPT (13,84,85),

and from a = 8 and b = 1 we get the PPT (20,21,29).

]]>x” + εx'(x^{2}-1)+x=0, where x=x(t) and ε>0.

If y^{2}(t)=tan^{2}(t)/(atan^{2}(t)+btan(t)+c), where a and c are >0, 0<b^{2}<4ac, and b<0,

and if we let z(t) be a ‘test’ function defined by z(t) =|y(t)| for 0<=t<=π

and z(t) = -|y(t)| for π<=t<=2π etc. then the 2π-periodic function z(t) for certain values of a, b, and c ‘seems’ to have the right kind of shape for the periodic solution to the Van der Pol equation – ignoring any scaling in the ‘t’ direction.

For example try a=0.28, b=-0.02 and c=0.004. This gives the following shape …

Note that the ‘first’ turning point in the first quadrant has t=tan ^{-1}(-2c/b) while the z coordinate is z(t)=2√c/(√(4ac-b^{2}).

If the amplitude is 2 then b^{2 }= c(4a-1). This requires that a>1/4.

Note also that if b=0 and a=c>0 then z(t) is just a sine curve.

The expression for y(t) suggests that the Van der Pol equation might be written as

x² = 1 -(x + x”)/(εx’) for the purposes of obtaining a sequence of approximations to x(t).

If in the RHS of x² = 1 – (x + x”)/(εx’) we let x = λsin(μt) then the LHS does not give an expression like that for y^{2}(t) as above.

However, if we re-write the Van der Pol equation as x² = (εxx’ – xx”)/(1 + εxx’)

and let x=λsin(μt) in the RHS, then the resulting expression,

(εxx’ – xx”)/(1 + εxx’) = (λ²μ²tan²(μt) + ελ²μtan(μt))/(tan²(μt) + ελ²μtan(μt) +1),

is very like that for y^{2}(t) as above.

If a = 1, b = 0.1 and c = 1 we get …

If a = 0.25, b = 0.2 and c = 0.05 we get …

If a = 100, b = 0.2 and c = 0.05 we get …

These shapes are very similar to those obtain by numerical solutions of the Van der Pol equation.

]]>Let the corresponding angle sizes be denoted by X, Y, and Z (assume radians).

Then X < π/2, since the triangle is acute-angled and so 2X < π = X+Y+Z. It follows then that X<Y+Z. Similarly it is easy to show that, Y<X+Z and Z<X+Y. So, X, Y and Z (the angle sizes of the acute-angled triangle) can also be the lengths of the sides of another triangle.

It should be fairly obvious that this only works if the original triangle is **acute-angled**, i.e. it does **not** work if the original triangle is right or obtuse.

By chance I happened to notice that for an acute triangle with angle sizes 45^{o} , 60^{o} and 75^{o} , the angle sizes also form a Pythagorean triple. Basically, these numbers are a multiple of (3, 4, 5). Any others?

The only other acute-angled triangles where this property holds are those where the angles are 30^{o} , 72^{o} and 78^{o} (5, 12, 13) and 18^{o} , 80^{o} and 82^{o} (9, 40,41).

The angle sizes of the original acute angled triangle can easily be shown to have a number of interesting and unexpected properties, such as …

3**YZ**/2π ≤ π/2 –**X**< 2**YZ/**π**,**

**XZ/**π < π/2 –**Y**< 2**XZ/**π, and

0 < π/2 –**Z ≤ 3****XY/**2π

**Proof** that 3**YZ**/2π ≤ π/2 –**X**< 2**YZ/**π.

Assume that (x,y,z) is **acute-angled** and that its corresponding angles are **X**,**Y**, and **Z**. Let the triangle **(X,Y,Z)** formed by (side lengths) **X**, **Y **and **Z **have corresponding angle sizes **A**, **B **and **C **as above.

Since we have assumed that 0<x≤y≤z, it follows that 0<**X≤****Y≤****Z** and so, since **Z**<π/2, **A**, **B** and **C** exist and 0<**A≤****B≤****C. **

We also have**, **π = **X**+**Y**+**Z **and π = **A**+**B**+**C**. Since (x,y,z) is acute angled, **X≤**π/3 and π/3≤**Z**<π/2. It must also be true that **A≤**π/3, and π/3≤**C**.

From 0<**A≤**π/3 we must have that 1/2≤cos(**A**)<1.

Hence 1/2≤(**Y**^{2}+**Z**^{2}–**X**^{2})/(2**YZ**)<1. So, **YZ≤****Y**^{2}+**Z**^{2}–**X**^{2}<2**YZ**.

Now π = **X**+**Y**+**Z **and so π –**X**=**Y**+**Z**.

Hence (squaring both sides) π^{2} – 2π**X**+**X**^{2} = **Y**^{2}+2**YZ**+**Z**^{2} or

π^{2}– 2π**X** – 2**YZ **= **Y**^{2 }+ **Z**^{2} – **X**^{2}. So,**YZ≤****Y**^{2}+**Z**^{2}–**X**^{2}<2**YZ** becomes

**YZ ≤**π^{2}– 2π**X** – 2**YZ** < 2**YZ** or 3**YZ ≤**π^{2}– 2π**X** < 4**YZ**.

**Finally**, dividing by 2π gives 3**YZ**/2π ≤ π/2 –**X**< 2**YZ/**π.

**Note: **If the triangle (**X**,**Y**,**Z**) is also acute angled, then the third set of inequalities above,

i.e. 0 < π/2 –**Z ≤**3**XY/**2π becomes **XY**/π < π/2 –**Z ≤**3**XY/**2π.

Also, if the triangle (**X**,**Y**,**Z**) is right angled then **XY**/π = π/2 –**Z. **This is easily proved and can be verified using some of the examples given above, e.g. if **X**=18^{o}, **Y**=80^{o }and **Z**=82^{o}. In fact, from **XY**/π = π/2 – **Z **it is easy to show that **Y = (**π/2-**X**)/(1-**X**/ π) or

**X = (**π/2-**Y**)/(1-**Y**/ π).

From **Y = (**π/2-**X**)/(1-**X**/π) and π =**X**+**Y**+**Z **we get **Z**= (π^{2}/2-π**X**+**X**^{2})/(π-**X**). Hence if (**X**,**Y**,**Z**) is right angled, then **Y** and **Z** are effectively determined/parametrised by **X**.

For example if **X** = π/6 then **Y = (**π/2-π/6)/(1-π/6/π) = π/3/5/6 = 2π/5 and **Z**= 13π/30. It is easily checked that π = π/6 +2π/5 +13π/30 and that …

**(13π/30) ^{2} = (π/6)^{2} + (2π/5)^{2}. **(

If (**X**,**Y**,**Z**) is right angled and isosceles then **X** = π/(2+√2), **Y**= π/(2+√2), and **Z**= π/(1+√2).

Now **X **lies in the range 0<**X≤**π/3 and as **X**-> 0, **Y **and **Z**-> π/2.

However, as **X**->π/3, **Y**-> π/4 and **Z**->5π/12, which is just a (3,4,5) triangle scaled by π/12. This can’t be the case though because we have assumed that x≤y≤z and so **X≤****Y≤****Z**.

**Y**>**X **while **X **is near 0 but **Y **decreases as **X **increases towards π/3 until **Y**=**X**= π(2-√2)/2. At that point (**X**,**Y**,**Z**) is right angled at **Z, **and **Z**= π(√2-1). If (**X**,**Y**,**Z**) is right-angled, and π(2-√2)/2<**X≤**π/3, it follows that **Y**<**X, **but this is not possible since we have assumed that **X≤****Y**.

If π(2-√2)/2<**X≤**π/3 it is straightforward to show that (**X**,**Y**,**Z**) is also acute angled.

First **π = X + Y +****Z **and **X≤Y **mean that **Z ≤****π**– 2**X** and since π(2-√2)/2<**X **it follows that **Z<**π(√2-1)<π/2. Hence the result.

Put another way, if you start with an acute-angled triangle where the smallest angle lies roughly from 52.7 degrees to 60 degrees and you generate another triangle using these angle sizes as the lengths of the sides, the generated triangle must also be acute-angled.

If however, 0<**X**≤π(2-√2)/2 then it is easy to show that the generated triangle can be acute, right or obtuse. **Note:** (2-√2)/2 ≈ 0.292893219.

As another example, if we take **X **= (1/4)π, and calculate **Y** using **Y = (**π/2-**X**)/(1-**X**/π), we get **Y **= (1/3)π, and then use π = **X**+**Y**+**Z**, we get **Z** = (5/12)π and of course …

((5/12)π)² = ((1/4)π)² + ((1/3)π)². (**Note**: (**X**, **Y**, **Z**) is just a (3,4,5) triangle).

If we extend/generalise this method to **X **= (p/q)π, where p and q are positive integers and

p/q < (2-√2)/2, and calculate **Y** and **Z** as above, then we arrive at the (familiar?) identity …

((q-p)² + p²)² = ((q-p)² – p²)² + (2p(q-p))² but having used little discernible number theory along the way.

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If the original acute-angled triangle with sides (x, y, z), assuming that x≤y≤z, and its ‘inside-out’ (**X**,**Y**,**Z**) are *similar *then …

x/**X**= y/**Y**= z/**Z**. But **X**+**Y**+**Z **= π. So, it follows easily that …

**X**= xπ/(x + y + z)

**Y**= yπ/(x + y + z) and

**Z**= zπ/(x + y + z)

Obviously an equilateral triangle is similar to its ‘inside-out’. I strongly suspect that this is the only acute angled triangle with this property.

**Proof**

Suppose we start with an **acute-angled **triangle where the sides are of length x, y, and z. Assume without loss of generality that 0 <x≤y≤z.

Let the corresponding angle sizes be denoted by **X**,**Y**, and **Z (assume radians)**.

Suppose also that (x, y, z) is similar to (**X**,**Y**,**Z). **Then(x, y, z) is **equiangular** to (**X**,**Y**,**Z).** By **equiangular** I mean that corresponding angles are equal. Also, corresponding sides are in the same ratio,so

x/**X** = y/**Y** = z/**Z**. Applying the sine rule to(**X**,**Y**,**Z) **it follows that **X**/sin(**X**) = **Y**/sin(**Y**) = **Z**/sin(**Z**).

The function f: (0,π/2) -> R defined by f(x) = x/sin(x), where x ∈ (0,π/2) is 1 – 1

and so if f(a) = f(b), a, b ∈ (0,π/2), then a = b.

It follows then that **X**/sin(**X**) = **Y**/sin(**Y**) = **Z**/sin(**Z)** implies that **X** = **Y** = **Z**, since** X**, **Y** and **Z**∈ (0,π/2), and so naturally x = y = z, i.e. (x, y, z) is **equilateral**.

**So if an acute angled triangle is similar to its ‘inside-out’ it can only be equilateral.**

——————————————-

**Note:** If (x,y,z) is an acute-angled triangle as above, with corresponding angles **X**, **Y** and **Z**, then (**X**,**Y**,**Z**) forms a triangle where **X** + **Y** + **Z** = π, or what might be called a **π-triangle**.

If (a,b,c) represents **any** triangle, then πa/(a+b+c), πb/(a+b+c), and πc/(a+b+c) represent the angles of an **acute angled** triangle.

Clearly πa/(a+b+c) + πb/(a+b+c) + πc/(a+b+c) = π and if (say) πa/(a+b+c)≥π/2 then a≥ b + c which contradicts the assumption that (a,b,c) forms a triangle.

—————————————–

Finally, a connection with **Fermat’s Last Theorem**.

If we let n be a whole number, n>=2, and assume that the (usual) equation,

z ^{n} = x ^{n} + y ^{n} , has a non-trivial solution in positive whole numbers x, y and z, then we can also assume without loss of generality that 0 < x < y < z. It is easy to show that, for such a non trivial solution, z < x + y, and so it follows that x, y, and z can form the sides of a triangle. In the case n=2, that triangle is right-angled, but in every other case, i.e. n>=3, it is straightforward to show that the triangle (x, y, z) is acute-angled.

**Proof – that (x, y, z) is acute-angled if n>=3.**

If n>=3, it is straightforward to show that z ^{2} < x ^{2} + y ^{2} if a non-trivial solution x, y, z to Fermat’s equation exists.

Let the corresponding angle sizes in triangle (x, y, z) be denoted by **X**,**Y**, and **Z.**

Then cos(**Z**) = (x^{2}+ y^{2}– z^{2})/(2xy), and so cos(**Z**)>0, i.e. **Z**(largest angle) is acute, so the triangle is acute angled.

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For a non-trivial solution to the Fermat equation,

z = cos(**Y**)x + cos(**X**)y and also z = (x/z)^{n-1}x + (y/z)^{n-1}y,

where 0<cos(**Y**)<1 and 0<(x/z)^{n-1}<1 etc.

Is it true then that cos(**Y**)= (x/z)^{n-1}and cos(**X**)=(y/z)^{n-1}??

In the case n=2 this is certainly true.

If either of these equations, cos(**Y**)= (x/z)^{n-1}or cos(**X**)=(y/z)^{n-1}, is true, then it follows that

z ^{n-2} = (y ^{n} – x ^{n} )/(y ^{2} – x ^{2} ).

If we assume that z ^{n-2} >= (y ^{n} – x ^{n} )/(y ^{2} – x ^{2} ) it is easy to show that a contradiction arises. Hence it is certainly true that z ^{n-2} < (y ^{n} – x ^{n} )/(y ^{2} – x ^{2} ).

**Proof**

If z ^{n} = x ^{n} + y ^{n} then z ^{i} < x ^{i} + y ^{i} is easily established for all i in the range 1, 2, … n-1.

Hence, z ^{n-2} < x ^{n-2} +y ^{n-2} .

So, z ^{n-2} >= (y ^{n} – x ^{n} )/(y ^{2} – x ^{2} ) implies that (x ^{n-2} + y ^{n-2} )(y ^{2} – x ^{2} ) > y ^{n} – x ^{n} .

Multiplying out this last inequality and simplifying gives x ^{n-4} >y ^{n-4} which is not possible if n>=4.

If n=3, we start with z>= (y ^{3} – x ^{3} )/(y ^{2} – x ^{2} ). Squaring both sides of this inequality, and using the fact that x ^{2} + y ^{2} > z ^{2} , gives after some rearrangement …

(y ^{4} – x ^{4} )(y ^{2} – x ^{2} ) > (y ^{3} – x ^{3} ) ^{2} . If this is multiplied out and simplified it leads to …

0 > x ^{2} – 2xy + y ^{2} or 0 > (x – y) ^{2} which again is not possible.

Hence the result is proved.

——————————————-

If n=3 this inequality becomes z<(y^{3}-x^{3})/(y^{2}-x^{2}) or z<(y^{2}+xy+x^{2})/(x+y). This last inequality leads to z-y<x^{2}/(x+y) and z-x<y^{2}/(x+y). It is straightforward to show that these inequalities are also true for all n>3.

If a non-trivial solution, (x,y,z), n>2, to the Fermat equation exists, then, as has been shown above, (x,y,z) is an acute angled triangle. An interesting construction can be drawn inside this triangle, which I would show here, if this system would only allow me to display it. Basically, a smaller (and similar) triangle (x(x+y-z)/z, y(x+y-z)/z,(x+y-z)) can be drawn inside (x,y,z).

Clearly since z^{n} = x^{n} + y^{n} then (x+y-z)^{n} = (x(x+y-z)/z)^{n} + (y(x+y-z)/z)^{n}. x+y-z is obviously an integer and it is easily shown that 0<x+y-z<z, 0<x(x+y-z)/z<x and 0<y(x+y-z)/z<y. If it can be shown that x(x+y)/z and y(x+y)/z are also integer then the method of Infinite Descent could be invoked.

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In the paragraphs immediately above, the terms x+y-z and xy/z appear. I wondered if there was some connection between these terms independent of Fermat’s equation. It turns out that there is, i.e. x+y-z<xy/z if z≠0 and (z-x)(z-y)>0.

**Proof**

Assume that z≠0 and (z-x)(z-y)>0. Consider xy/z-(x+y-z).

xy/z-(x+y-z) = (xy-z(x+y)+z^{2})/z = (x-z)(y-z)/z = (z-x)(z-y)/z>0. Hence the result is true.

So if we had (say) 0<x<y<z then (using explicit * for multiplication and / for division) we have **x+y-z<x*y/z**. Clearly the LHS and RHS of < have the same structure but just different operators inserted.

There are other similar inequalities e.g. if 0<x<y<z, then** x-y+z>xz/y**. If we consider the expression x-y+z-xz/y then this is equal to …

-(1/y)(y^{2} – y(x+z) + xz) = -(1/y)(y-x)(y-z) = (1/y)(y-x)(z-y)>0. Hence this result is also true.

Again if 0<x<y<z, then **-x+y+z<yz/x**.

Consider -x+y+z-yz/x = (-1/x)(x^{2} -x(y+z) +yz).

Then,-x+y+z-yz/x = (-1/x)(x-y)(x-z) =(-1/x)(y-x)(z-x)<0. Hence the result.

]]>If (x,y,z) and (a,b,c) T then it is easy to show that if + is defined in the obvious way, i.e. (x,y,z) + (a,b,c) = (x+a,y+b,z+c) then (x,y,z) + (a,b,c) T. + is commutative and associative.

Clearly if k>0, and k(x,y,z) is defined by (kx,ky,kz) then k(x,y,z) T.

A type of multiplication * can also be defined, e.g.

(x,y,z)*(a,b,c) = (xa+yc+zb, yb+xc+za, zc+xb+ya)

This * operation is closed on T, commutative, distributive over addition, but not associative.

If we define the perimeter function p:T—>R ^{+} in the obvious way, i.e. p((a,b,c))=a+b+c, then if t1, t2, and t3T , p(t1+t2) = p(t1)+p(t2), p(kt1) = kp(t1), and p(t1*t2) = p(t1)*p(t2)

Even although (t1*t2)*t3 ≠ t1*(t2*t3) the perimeter function makes

p((t1*t2)*t3)=p(t1*(t2*t3)).

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If (x,y,z) T is an acute angled triangle with corresponding angles **X**,**Y**,**Z**, then it is easy to show that (x^{2},y^{2},z^{2}) and (**X**,**Y**,**Z**) also belong to T. This is not true for obtuse or right triangles. For a proof that (**X**,**Y**,**Z**) belongs to T if (x,y,z) T is an acute angled triangle, see

‘A curious property of Acute Triangles’.

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If (x,y,z) T then (x ^{1/n} , y ^{1/n} ,z ^{1/n} ) T for any integer n>=2 and each of these triangles is also acute. Clearly, since x ^{1/n} -> 1 for any x>0 as n -> ∞, (x ^{1/n} , y ^{1/n} , z ^{1/n} ) -> (1,1,1)

as n -> ∞.

———————–

Now, suppose (a,b,c) T with a<b<c. Suppose also that (a,b,c)+(c,a,b) is right angled. Then (a,b,c)+(c,a,b) =(a+c,b+a,c+b), where b+a<c+a<c+b.

If (a+c,b+a,c+b) is right-angled then (c+b) ^{2} = (b+a) ^{2} + (a+c) ^{2} from which it follows that

c = a(a+b)/(b-a). Since c>b then it follows that a(a+b)>b(b-a) and so (a+b) ^{2} > 2b ^{2.}

Hence a>(√ 2 – 1)b, and c<a+b means that a<b/2.

Using the above we can easily generate some Heronian triangles.

Choose a value for b such as 17. Then (√ 2-1) x 17<a<17/2 or 7.04163056<a<8.5.

Take a=8. Then c=200/9 or 22 2/9. Hence, if b=17 and a=8,

(a+c, b+a, c+b) = (30 2/9, 25, 39 2/9) or 1/9(272, 225, 353) which is Heronian.

———————–

Now suppose (a,b,c) is scalene i.e.a<b<c for example. Can (a,b,c) be written as a sum of 2 isosceles triangles? Suppose so. Then, assume that (a,b,c)=(x,x,y)+(p,q,q). It follows that a=x+p, b=x+q, and c=y+q. From these equations then q-p=b-a>0 and y-x=c-b>0.

In (x,x,y) then x<y<2x, or 0<y-x<x. From (p,q,q) we get no additional information. So, given (a,b,c) with a<b<c, choose x so that c-b=y-x<x<a. Then choose y=x+(c-b). p then follows from p=a-x and q=p+(b-a).

Now suppose we apply this to (5,12,13) then y-x=1 and q-p=7. So, 1<x<5. If x=3 then y=4. Hence, p=2 and q=9. So, (5,12,13)=(3,3,4)+(2,9,9). Clearly this is not the only way this can be done. E.g. (5,12,13)=(2,2,3)+(3,10,10)=(4,4,5)+(1,8,8) and these are just integer side lengths.

**Note:**the 2 isosceles triangles mentioned here are fundamentally different since (x,x,y), x<y, can be acute, right or obtuse, while (p,q,q), p<q, can only be acute. Perhaps (x,x,y) should be called type i1 and (p,q,q) type i2. Hence, all triangles are either e(equilateral), i1, i2 or s(scalene) = i1+i2.

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If we start with 2 positive numbers x and y, and assume that x<y<2x, we can generate 2 isosceles triangles from these immediately, i.e. (x,y,y) and (x,x,y). If we add these together we get … (x,y,y) + (x,x,y) = (2x,x+y,2y). Clearly, 2x<x+y<2y so (2x,x+y,2y) is scalene. Right angled?? If so then (2x) ^{2} + (x+y) ^{2} = (2y) ^{2} .

From that it follows that 3y ^{2} – 2xy -5x ^{2} = 0.

Solving this equation for y in terms of x we get y = 5x/3 or -x. So, y= 5x/3. Hence (2x,x+y,2y) = (2x,8x/3,10x/3) = 2x/3(3,4,5) or a scaled version of a (3,4,5) triangle.

Interestingly, if we try to ‘multiply’ the triangles (x,y,y) and (x,x,y) using the * operation we get (x ^{2} + xy + y ^{2}, 3xy, x ^{2} + xy + y ^{2} ) which is also isosceles but not right-angled.

———————–

**Acute angled triangles have some interesting properties.** If we add in the conditions that such a triangle is also integer, i.e. has integer sides, and is scalene, then only certain lengths of side are possible.

Suppose a scalene, integer, acute angled triangle has sides (x,y,z) with 0<x<y<z.

**What is the ‘smallest’ such triangle?**

Clearly, 1≤x, 2≤y, and 3≤z and also z^{2}<x^{2}+y^{2}.

Then (z – y + y)^{2}<x^{2} + y^{2 }and so (z-y)^{2}+2y(z-y)<x^{2}. From this (last) inequality it follows that x^{2}>5 and so x≥3. x≥3 implies that y≥4 and so from the same inequality it follows that x^{2}>9, or that x≥4.

If x=4, then y≥5 and so, using z^{2}<x^{2}+y^{2}, it follows easily that z-y<x^{2}/(z+y)≤16/11. Hence, z-y≤1. But z-y>=1, and so it must be that z-y=1. From this it follows that 1+2y<16 or y<=7. So y=5, 6, or 7. Hence the ‘smallest’ scalene, integer, acute triangles are (4,5,6), ( 4,6,7) and (4,7,8).

Note that the ‘largest’ of these is very nearly right-angled since 8² = 4² + 7² – 1.

———————–

So, if (x,y,z) is a scalene, integer, acute-angled triangle then, assuming that x<y<z, we can also assume that x>=4. Also, z^{2}<x^{2}+y^{2} since the triangle is acute-angled.

Then (z – y + y)^{2}<x^{2} + y^{2} and so (z-y)^{2}+2y(z-y)<x^{2}.

Now x+1 ≤ y so it follows that (z – y + x + 1)² < x² + (x+1)² .

Hence, z – y + x + 1 ≤ [√(x² + (x+1)²)] and so z – y ≤ [√(x² + (x+1)²)] – (x+1).

(z-y)^{2}+2y(z-y)<x^{2 }in turn implies that y<(x^{2} -(z-y)^{2})/2(z-y).

So, to generate such triangles, first choose **x>=4**.

Then choose **z-y** so that 1 ≤ z – y ≤ [√(x² + (x+1)²)] – (x+1).

Now choose **y** so that x+1 ≤ y < (x^{2}-(z-y)^{2})/2(z-y).

z is of course z-y + y.

Suppose, for example, that x=10, then 1≤z-y≤3, or z-y is 1, 2 or 3.

If z-y=1 then 11≤y≤49. The triangles are therefore (10, 11, 12), (10, 12,13) … (10, 49, 50).

If z-y=2 then 11≤y≤23. The triangles are (10, 11, 13), (10, 12, 14) … (10, 23, 25).

If z-y=3 then 11≤y≤15. The triangles are therefore (10, 11, 14), (10, 12, 15) … (10, 15, 18).

In all such triangles we have z<x+y, so z≤x+y-1 since x, y and z are integers.

Suppose there is a case where z=x+y-1. Then since z^{2}<x^{2}+y^{2} it follows that (x+y-1)² < x² + y²

or, after some simplification, 2(xy – x – y)<-1.

From this point it can be shown easily that x<2, which is not possible since, clearly, x≥4.

Hence, z<x+y-1.

If we repeat this argument with z=x+y-2, then it follows, similarly, that z<x+y-2.

If we try to push this further to z<x+y-3 we get x=4 and z=y+1, which is the case for the ‘smallest’ scalene, integer, acute triangles (4,5,6), ( 4,6,7) and (4,7,8).

**So, for a scalene, integer, acute triangle (x,y,z), where x<y<z, we must have x≥4 and z<x+y-2.**

———————–

If (a,b,c) is an acute angled triangle, a≤b≤c, and (d,e,f) is also acute angled, d≤e≤f, is it true that (a+d,b+e,c+f) is also acute angled?

**Note**: while this question initially seemed to be answered with a ‘yes’ it is in fact ‘no’.

**Proof**

If 0<x<1, then (3, 4, 5-x) and (5, 12, 13-x) are both **acute** angled but their sum

(3,4,5-x) +(5,12,13-x) or (8,16,18-2x) is easily shown to be obtuse if x < 9 -4√5 = 0.05572809.

———————–

However, if (a,b,c) is a triangle, a≤b≤c, but not acute angled, and (d,e,f) is also not acute angled, d≤e≤f, then(a+d,b+e,c+f) is also not acute angled.

Since (a,b,c) and (d,e,f) are not acute angled then a² + b² ≤ c² and d² + e² ≤ f².

We need to prove that (a+d)² + (b+e)² ≤ (c+f)² to show that (a+d, b+e, c+f) is not acute angled. It follows easily that (a+d)² + (b+e)² – (c+f)² = a² + b² – c² + d² + e² – f² +2ad + 2be – 2cf

and so (a+d)² + (b+e)² – (c+f)² ≤ 2ad + 2be – 2cf.

Suppose ad + be – cf >0. Then (ad+be)² >c²f² ≥ (a²+b²)(d²+e²) .

So, 2adbe > a²e² + b²e² or 0 > (ae – bd)² which is clearly not true.

Hence the result is proved.

———————–

]]>Cubes can also be represented as differences of squares, e.g. n^{3} = (n(n+1)/2)^{2} – (n(n-1)/2)^{2}.

An application of the above identity is to∑_{1≤k≤n}k^{3} = (n(n+1)/2)^{2}.

For higher powers, if we try n^{a} = (n^{b}(n^{c} + 1)/2)^{2} – (n^{b}(n^{c}-1)/2)^{2} ,

where a, b and c are whole numbers, then a=2b+c. So, if a=5, we can have b=2 and c=1 or b=1 and c=3.

Hence, for n^{5}(say) we can have either n^{5} = (n^{2}(n+1)/2)^{2} – (n^{2}(n-1)/2)^{2 }or

n^{5}= (n(n^{3}+1)/2)^{2}– (n(n^{3}-1)/2)^{2}.

**Divisibility by 3**

If (x,y,z) is a PPT (Primitive Pythagorean Triple) then z^{2} = x^{2} + y^{2}. Looking at examples of PPT’s it is evident that **exactly** one of x and y is divisible by 3. This follows from an application of n^{3}≡ n mod 3 for any integer n.

**Proof**

If z^{2}= x^{2}+ y^{2 }^{}then 2x^{2}y^{2} = z^{4} – x^{4} – y^{4 }≡ z^{2}– x^{2}– y^{2 }mod 3≡ 0 mod 3 and so 3|2x^{2}y^{2}. Hence, 3 divides x or y or both. 3 can’t divide both x **and** y because then 3 would then divide z, since z^{2}= x^{2}+ y^{2}, meaning that (x,y,z) is not primitive. Hence the result.

For positive integer powers of positive integers it is clear that these can be expressed as differences of squares of positive integers. What is also clear is that in all cases at least one of these squares must be divisible by 3.

**Case 1 Odd powers, 3 and higher**

n^{2k+1} = (n^{k}(n+1)/2)^{2} – (n^{k}(n-1)/2)^{2 }, k≥1

**Case 2 Even powers, 4 and higher**

n^{2k} = (n^{k-1}(n^{2} + 1)/2)^{2} – (n^{k-1}(n^{2} – 1)/2)^{2 }, k≥2

Obviously if 3|n then both squares in the above cases are divisible by 3. Otherwise, in **Case 1** either the first or the second square must be divisible by 3 depending on n, while in **Case 2** the second square is always divisible by 3.

**Divisibility by 5**

If (x,y,z) is a PPT (Primitive Pythagorean Triple) then z^{2}= x^{2}+ y^{2}. Looking at examples of PPT’s it is evident that **exactly **one of x, y and z is divisible by 5. This follows from an application of n^{5}≡ n mod 5 for any integer n.

**Proof**

If z^{2}= x^{2}+ y^{2 }^{}then z^{6} = x^{6} + 3x^{4}y^{2} + 3x^{2}y^{4} + y^{6} = x^{6} + y^{6} + 3x^{2}y^{2}z^{2}.

Hence, 3x^{2}y^{2}z^{2} = z^{6} – x^{6} – y^{6}≡ z^{2} – x^{2} – y^{2} mod 5≡ 0 mod 5. Therefore 5|3x^{2}y^{2}z^{2} and so **the result follows**.

If we look at n^{2k}= (n^{k-1}(n^{2}+ 1)/2)^{2}– (n^{k-1}(n^{2}– 1)/2)^{2}, k≥2, then the product of the terms to be squared gives n^{2k-2}(n^{4} – 1)/4 = n^{2k-3}n(n^{4}– 1)/4≡ 0 mod 5, so at least one of the squared terms must be divisible by 5.