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Prove that $x^2 + y^2 = z^n$ has a solution in $\mathbb{N}$, for all $n$ belonging to the set of natural numbers.

I think it uses Induction as a method of proof but just can't proceed with three variables. Any help is greatly appreciated.

nonuser
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Anonymous
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  • If $0\in\mathbb{N}$ (as it is in many conventions for the natural numbers), then $(x,y,z)=(0,0,0)$ is always a solution. – Barry Cipra Nov 09 '18 at 19:58
  • Hint: $1^2 + 0^2 = 5^0$ and $x^2 + y^2 = 5^n \implies (2x-y)^2 + (2y+x)^2 = 5^{n+1}$ – achille hui Nov 09 '18 at 20:09
  • @BarryCipra: I don' agree. For such polynomial equations, allowing zeroes corresponds to a simpler problem, with less terms, which can be of no interest. For instance, young Fermat tackled the equation $x^n+0^n=z^n$. –  Nov 09 '18 at 20:14
  • $|(a+ib)^n|^2 = (|a+ib|^2)^n$ , where $i$ is the imaginary unit. – user90369 Nov 09 '18 at 21:00
  • @Winter, lulu, Dietrich Burde though I see that the problem exists but the question is different from the one you have given reference to in the matter of approach. Here the problem statement was required to be solved using Induction only but the problem to which you have given reference to does not use Induction. So, this question was not only about proving the statement but rather find a way to apply Induction. It would be polite of you if could remove the "Marked As Duplicate" tag from this question. – Anonymous Nov 10 '18 at 04:28

1 Answers1

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Proof by induction:

base case: $n = 1$

$x^2 + y^2 = z$

for any integers $x,y, x^2 + y^2$ is an integer

Suppose for all $k\le n$ there exists integers $x,y,z$ such that

$a^2 + b^2 = z^k$ and $p^2 + q^2 = z$

We must show that there exists integers $x,y,z$ such that:

$x^2 + y^2 = z^{n+1}$

$z^{n+1} = z(z^n) = (p^2 + q^2)(a^2 + b^2)\\ (ap)^2 + (aq)^2 + (bp)^2 + (bq)^2\\ (ap)^2 + (aq)^2 + (bp)^2 + (bq)^2 + 2abpq - 2abpq\\ (ap + bq)^2 + (bp - aq)^2 = z^{n+1}$

Doug M
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  • can you please explain the last part of the proof where you prove the existence of a solution for n+1 ? – Anonymous Nov 10 '18 at 07:12
  • If you choose any two integers $a,b, z=a^2 + b^2$, then $(a^2 - b^2)^2 + (2ab)^2 = z^2$. This process can be repeated indefinately. – Doug M Nov 12 '18 at 02:22