Given the equation: $$x^2 + y^2 = z^n $$ Prove that there are solutions in $\mathbb N$. Provided that $n\in\mathbb N.$
Please don't give me the full answer, rather a way to attack the problem, I have no idea how to tackle this proof.
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1Some people include $0$ in $\mathbb N$, others do not. Is zero a natural number? – Thomas Andrews Sep 29 '17 at 04:52
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And we have no idea what answer to give because we don't have any context. I can think of a theorem that solves your problem, but you may not be that far into the book yet. – B. Goddard Sep 29 '17 at 04:53
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1Do you know about the Gaussian integers, $\mathbb Z[i]$? – Thomas Andrews Sep 29 '17 at 04:55
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1If you want to prove the existence of a solution, can't you just show a solution that works? e.g. $$3^2+4^2=5^2$$ where $z=5$ and $n=2$ – Sep 29 '17 at 04:56
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I think the point is to show that is a solution for each $n$. @Jazzachi – Thomas Andrews Sep 29 '17 at 04:56
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@ThomasAndrews Ah okay. – Sep 29 '17 at 04:57
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@B.Goddard It's in the second chapter of "challenge and thrill of pre college mathematics" called "the arithmetic of integers" – user167920 Sep 29 '17 at 04:59
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Also a duplicate of [this](https://math.stackexchange.com/questions/1248042/prove-that-x2y2-zn-has-a-solution-in-mathbbn-for-all-n-in-mathb – Ross Millikan Sep 29 '17 at 05:12
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I just streched my arm to my book shelf and took the book challenge and thrill of pre college mathematics. But sadly, it was a duplicate. – Vidyanshu Mishra Sep 29 '17 at 05:17