1
$\begingroup$

The problem is prove $x^3 + y^3 = z^3$ has no positive integer solutions. I understand this was proven in the 1700s by Euler, but I cannot find any books or in-depth references on the proof.

Here are some proofs I have tried to follow:

https://www.mathpages.com/home/kmath009/kmath009.htm

http://www2.washjeff.edu/users/mwoltermann/Dorrie/21.pdf

Reference request: Clean proof of Fermat's last theorem for $n=3$.

The equation $x^3 + y^3 = z^3$ has no integer solutions - A short proof

In your professional opinion, what would be the first step to understanding the proofs linked above? Do you know of any books or other references that go into more detail on this problem?

Also this is one attempt I made:

Starting with assuming $x^3+y^3=z^3$ has a solution such that $xyz\neq0$ and x,y,z are co-prime.

$\therefore z^3-x^3-y^3 + (x+y-z)^3 = (x+y-z)^3$

I believe this simplifies to:

$3(x+y)(z-x)(z-y)=(x+y-z)^3$

But aren't these factors also co-prime and therefore not cubic?

$\endgroup$
  • 2
    $\begingroup$ What is wrong about the proofs given at your links? The first one, for example, is really very detailed. I would try to follow it. For shorter proofs you will definitely need more algebraic number theory, or at least some commutative algebra. $\endgroup$ – Dietrich Burde Oct 27 '20 at 15:20
  • 3
    $\begingroup$ another reference would be 13 Lectures on Fermat's Last Theorem by Paulo Ribenboim, Lecture III Section 3 "The Cubic Equation" $\endgroup$ – J. W. Tanner Oct 27 '20 at 15:24
  • $\begingroup$ I get stuck when they rush over some explanations in the lemmas. I was just wondering if there are any other references. $\endgroup$ – Joseph Oct 27 '20 at 15:24
  • $\begingroup$ Thank you for that reference. $\endgroup$ – Joseph Oct 27 '20 at 15:31
  • $\begingroup$ For example, many of these proofs make statements like the following: (1) Let x be a positive, odd factor of a2 + 3b2 (2) Then, there exists a value f such that: a2 + 3b2 = xf (3) We can assume that x is greater than 1 (since if x = 1 it's only factor is itself and 1 = 12 + 3(0)2.) (4) There exists integers m,n such that: a = mx ± c b = nx ± d where c,d can be positive or negative. That is all fine until step 4. I do not understand why a = mx+c, b=nx+d. To me it is not obvious. $\endgroup$ – Joseph Oct 28 '20 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.