Prove that for any $n\in N, n\neq0$, there exist $x,y,z\in N$, all non-zero, such that $x^n =y^{n-1}+z^{n+1}$.
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2There exists a non-zero solution for all $n>0$. Hint: look for solutions where $y^{n-1} = z^{n+1}$. What do $y$ and $z$ need to look like for that to happen? – Erick Wong Sep 29 '16 at 18:21
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@Avi: To many people $0\in\Bbb{N}$. – Jyrki Lahtonen Sep 29 '16 at 18:22
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1Is zero a natural number? http://math.stackexchange.com/questions/283/is-0-a-natural-number – Alex Silva Sep 29 '16 at 18:25
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1$2^2=1^3+3^1$ How does this sound? – Sep 29 '16 at 18:26
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Claudiu, I edited the question the best I could (with a bit of educated guessing). Please comment if it still looks strange :-) – Jyrki Lahtonen Sep 29 '16 at 18:27
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1And now you can start thinking in terms of Erick Wong's nice hint! I will purge the early comments as obsolete. – Jyrki Lahtonen Sep 29 '16 at 18:28
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@AlexSilva Thanks for the reference. ( I grew up believing natural numbers start from 1, so it was interesting to see that post) – Sep 29 '16 at 18:29
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@Avi, you're welcome! – Alex Silva Sep 29 '16 at 18:30