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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