Mythic Algebra and Other Things

Euclid and Fermat

By Michael Griffin MLS

This is the second article about using ancient Greek formulas with Fermat's Last Theorem.

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 in the power of two:

(m2-n2)2 + (2mn)2 = (m2+n2)2 is A2 + B2 = C2 and we get:

m4-2m2n2+n4 + 4m2n2 = m4+2m2n2+n4

Cancelling terms leaves -2m2n2 + 4m2n2 = 2m2n2 or 4m2n2 = 4m2n2

A trivial result that we see has any possible solutions for m and n.

When we go to higher powers we get instead binomial expansions, with alternating +/- terms on the left, and only positive terms canceling on both sides of the equation.

If N=3

(m2-n2)3 + (2mn)3 = (m2+n2)3 is A3 + B3 = C3 and we get:

(m2-n2)( m4-2m2n2+n4 ) + 8m3n3= (m2+n2) (m4+2m2n2+n4 ) then distributing:

m6-2m4n2+m2n4 - n2m4+2m2n4 - n6 + 8m3n3 = m6+2m4n2+m2n4+n2m4+2m2n4+n6

Cancelling identical terms on both sides leaves:

-2m4n2 - n2m4 - n6 + 8m3n3 = 2m4n2+n2m4+ n6

Except for the term 8m3n3 everything on the left is just the negative of what is on the right so let us combine like terms:

8m3n3 -4m4n2 -2n2m4 -2n6 = 0 and then we can divide all by 2

4m3n3 -2m4n2-n2m4 -n6 = 0

then what remains after expansion and cancellation is:

4m3n3 - 3m4n2 - n6 = 0 and m=n is the solution of this. That is not allowed since the only resulting triple would be (0, 2m2, 2m2) which also shows why we need m>n.

N = 3 is the first case where N is odd. What about when N is even? Let's examine the expansion of the A term with its subtraction components

if N = 4:

(m2-n2)4 = (m2-n2)2(m2-n2)2 = ( m4-2m2n2+n4 )( m4-2m2n2+n4 ) =

m8-2m6n2+m4n4 -2m6n2+4m4n4 -2m2n6+ m4n4 -2m2n6+n8

The positive terms would be cancelled with their matches in the C term on the other side of the equation, leaving the negative terms: -4m6n2 -4m2n6

If we bring the remaining C terms over to the left side and write out the full equation we now have:

(2mn)4 -8m6n2 -8m2n6 = 0 Then dividing by 8:

2m4n4 -m6n2 -m2n6 = 0 And once again m=n is the natural number solution.

In general, a binomial expansion and cancellation leaves:

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

By intuition, any higher powers would result in the same balancing of +/- terms and a final result with m=n the only solution. So if N>2 there is no m>n solution to Euclid’s formula.

∎

QED

Note, it is an unfounded assumption that just because Euclid’s formula applies to N=2 it should generalize to all powers. What this exercise shows is that it cannot.

Sources

Fermat's Last Theorem

From Wikipedia, the free encyclopedia

This page was last edited on 31 August 2020

https://en.wikipedia.org/wiki/Fermat's_Last_Theorem

Formulas for generating Pythagorean triples

From Wikipedia, the free encyclopedia

This page was last edited on 24 June 2020

https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

Proof of Fermat's Last Theorem for specific exponents

From Wikipedia, the free encyclopedia

This page was last edited on 19 October 2020