Mythic Algebra and Other Things: September 2024

Euclid and Fermat again

By Michael Griffin, MLS

This is the fourth article about using ancient Greek formulas with Fermat's Last Theorem. It begins with parts of the previous article about Euclid's formula:

In Fermat's problem:

AN + BN = CN has infinitely many solutions for N but the highest whole number solution is 2, as with the sides of a right triangle in the Pythagorean theorem: A2 + B2 = C2

As with Fermat’s theorem, this is easily true if we make no restrictions on the values of A,B, or C. For the sake of argument and the rules of the game, A,B,C, and N in Fermat’s, are natural numbers only.

Euclid is credited with a formula to generate all possible Pythagorean triples: for any natural numbers m and n, m>n, a triplet of (A,B,C) results from

(m2-n2, 2mn, m2+n2).

For example, (m,n) of (2,1) makes (3,4,5), the first such triple, for 9+16=25.

Let us begin by expanding Euclid’s formula with an unfounded assumption that just because Euclid’s formula applies to N=2 it should generalize to all powers. What this exercise will show is that it cannot.

So for AN + BN = CN we use (mN-nN)N + (2mn)N = (mN+nN)N.

First, to see that AN + BN = CN has infinitely many solutions for N, look at the drawing below. The height of the curve CN can be arbitrarily divided into any two parts that will satisfy AN + BN.

The question is now are there any values for m,n that will satisfy:

AN = (mN-nN)N and BN = (2mn)N and CN = (mN+nN)N.

The previous article showed that if we do a binomial expansion for any power of N and then cancel like terms on both sides of the equation what we will have remaining is

BN = (2mn)N = 2(+/- leftover terms) and so m=n is the solution.

Putting this solution into Euclid's formula we see:

AN = (mN-mN)N = 0 and BN = (2mm)N = (2m2)N and CN = (mN+mN)N = (2mN)N

So 0 + BN = CN or (2m2)N = (2mN)N so 2Nm2N = 2N(mN)N

Cancelling the 2N on both sides leaves m2N =(mN)N

And the only solution for the exponent is N = 2

So we see that altering Euclid's formula does not change the result when we still used the only solution of an invalid triple where the A term is zero.

We make an intuitive leap to conclude that if the Greek formulas comply with Fermat's, then the solution sets of these formulas will always comply also. While these investigations may then have established Fermat's Last Theorem for all Pythagorean triples, that leaves out all other possible combinations of numbers, that is all ABC’s that do not fit Euclid's formula. These would differ from Euclid's values as:

A = m2-n2 plus or minus 0, 1, 2, ...

B = 2mn plus or minus 0, 1, 2, ...

C = m2+n2 plus or minus 0, 1, 2, ...

If we kept the same factoring of these according to Euclid's formula there would be extra bits resulting in

BN = (2mn)N = 2(+/- leftover terms) plus or minus 1, 2, ... and so m=n is NOT the solution. Also since the ABC’s do not fit Euclid's formula there is no solution for N = 2.

We'd be starting all over, with an infinite endless supply of unproved triples, minus the set of Pythagorean triples.

Does Euclid's formula cover all possible solutions? It does generate all Pythagorean triples if a K factor is put onto its formula for ABC. For example if K is equal to 2 then (3,4,5) and (5,12,13) become (6,8,10) and (10,24,26). And if we interpret Euclid's derivation of the formula as a proof of all possible solutions for N=2 then in fact to prove Fermat's is it sufficient to show that Euclid's has no solutions beyond N=2? That has now been shown.

It would be sufficient if it could be shown that all solutions of AN + BN = CN

must comply with Euclid's. That is easy for many if not all triples, so it is not enough. We can only pursue it as another possible proof just on the set of Pythagorean triples.

In general, if A,B,C,N,k,m,n are real numbers then m and n can always be chosen such that the Nth root C = N√(m2-n2)N + (2mn)N. There are infinitely many solutions in real numbers, just not integers or natural numbers. That is not necessary to comply with Euclid's formula. Restricting these to natural numbers results in Pythagorean triples. If N=2 the result for the Nth root C is Euclid’s definition for the C term, m2+n2. If N is beyond 2 we do not get Euclid’s C term and Fermat's Last Theorem means at least one of A,B,C is not an integer. The next post will discuss that.