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My proof goes like this

For $n=2$ there are infinite solutions.

Let $x^2+y^2=p^2$ be one such solution.As there are infinite solutions, We can assume that $ p = q^3$ for some integer $q$.

Then after substituting,we get

$x^2+y^2=(q^3)^2$

$x^2+y^2=(q^2)^3$

We found a solution for $n=3$.Instead of $p=q^3$ we can put any power to q.Therefore we can find a solution for any value of $n$.

Is this correct proof or are there any loopholes in my proof or is there any more elegant proof

Mathematical Curiosity
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    The proof is wrong. Infinite solutions does not mean that there is a solution with $p=q^n$ for some $q$. For instance consider the equation $a+2=b^3$ in $\mathbb N$> Since the difference between cubes is at least $7$ in $\mathbb N$ there will be not solutions where $a$ is a cube. However there are infinite solutions $a=t^3-2$, $b=t$ $t>1$ – Ben Martin Oct 03 '19 at 06:41
  • I mean, if we think x,y,z as sides of a right angled triangle then z side can be perfect cube or perfect nth number – Mathematical Curiosity Oct 03 '19 at 06:46
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    @HVxvejjw The statement you wrote is false. You can't express $21$ as a sum of two squares. – Wojowu Oct 03 '19 at 09:17
  • @Wojowu, you are right. It's partially true. I should have thought a bit more before writing it. –  Oct 03 '19 at 09:21

1 Answers1

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As there are infinite solutions, We can assume that $ p = q^3$ for some integer $q$.

This is wrong. You can assume that there's a solution with $p > k$ for any finite $k$, but that's about all you get from the infinitude.

Peter Taylor
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