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For pythagorean triples we have many examples when in the set of these three numbers two of them differ only by 1.

For example $(5^2+12^2=13^2)$, $(7^2+24^2=25^2),(9^2+ 40^2 = 41^2),(11^2 + 60^2= 61^2)$ etc..

Is it known the proof that there is no such triples for integers $x,y,z= y+1$ and $n>2$ to satisfy $x^n+y^n=(y+1)^n$ ?

I know that this is subcase for Fermat's Last Theorem which has now a general proof but it is quite complicated.

I would like to know a relatively simple proof for this seemingly much simpler subcase for FLT with constraint for integer triples mentioned above.

Edit

If general proof is impossible to find please at least show no existence of solution for case $n=3$. In this case $x^3=(y+1)^3-y^3= 3y^2+3y+1$ which leads to $\dfrac{x^3-1}{3}=y(y+1)$. There is no integers satisfying this equation?

Widawensen
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  • A few initial observations: * one needs to exclude $x=1$ and $y=0,-1$; * it suffices to prove this for odd primes $n$ and for $n=4$; * when $n$ is prime, any solution satisfies $x\equiv1$ (mod $n$) by Fermat's little theorem. – Greg Martin Oct 23 '22 at 18:10
  • See also this post, or this post. – Dietrich Burde Oct 23 '22 at 18:16
  • @DietrichBurde Thank you for these two examples. BTW the question from 2015 became duplicate of the question from 2018 :) How is it possible ? – Widawensen Oct 23 '22 at 19:14
  • The $2015$ post has no answer at all. So one cannot refer to it an an answered question. So it is not possible to use it for a duplicate. In general, I think, we only need one post with a complete answer (not necessarily the earlier one, if two exist in parallel). So all other ones become a "duplicate" (may be you had another idea what duplicate means, retaliated to the year). – Dietrich Burde Oct 23 '22 at 19:15
  • I was assuming that the later one is the duplicate of an earlier one .. – Widawensen Oct 23 '22 at 19:26
  • But if the earlier one is just a question without answer, it will not answer future identical posts. – Dietrich Burde Oct 24 '22 at 08:40
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    I am not an expert in such diophantine equations , but that it only contains two unknowns should make a proof significantly easier. – Peter Oct 24 '22 at 09:43
  • @Peter I was assuming that exactly it would be much easier. But surprisingly, from the marked answer of twin linked question "This is an open question". It was stated in 2018, maybe today it is also true. – Widawensen Oct 25 '22 at 10:41

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