Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can we reduce Fermat last theorem problem to the case $z=x+1$ where $x^n + y^n = z^n$?

Why am I asking that? I found in that case and in case $n=3$ that difference of cubes: $1$,$7$,$19$,$37$

is a combination of diferrence of next sequence:

share|cite|improve this question
Prior to Wiles's theorem that there is no integer solution, there was no such reduction, and Wiles's proof does not go through such a reduction. – André Nicolas Apr 8 '12 at 20:17
I think we may be able to reduce to the cases $n=1$ or $n=2$. The condition specified adds nothing in either of these cases. – Mark Bennet Apr 8 '12 at 20:20
I don't get what you mean: Centered hexagonal numbers are combinations of differences of $\sum (-1)^{n+1} n^3$? – draks ... Apr 8 '12 at 20:33
1-0=1,7-0=7,20-1=19,44-7=37,81-20=61,135-44=91 and so on.... – Bojan Vasiljević Apr 8 '12 at 20:43

What has long been known (and proved by elementary methods) , in the crucial case that $n$ is prime and $x,y$ and $z$ are pairwise coprime, is that in the case ${\rm gcd}(n,xyz) =1,$ we have $z-x = u^{n}, z-y = v^{n},$ and $x+y = w^{n},$ for integers $u,v$ and $w.$ In the case that $n$ is prime and $n$ divide $xyz$ (but $x,y$ and $z$ still pairwise relatively prime) we either have $z-x = n^{n-1}u^{n}$ or $z-x = u^{n}$ for some integer $u.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.