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

**Euclid’s formulae.**

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).

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**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.

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**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).

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