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True of False: If $m$ and $n$ are odd positive integers, then $n^2+m^2$ is not a perfect square.

Anyway it is already appear here,but I want check my solution!

The statement is true, because , suppose $$n^2+m^2=k^2$$ Then $n^2=k^2-m^2=(k-m)(k+m)$. Here divisors of $n^2$ are $1,n,n^2$, so either

  • $k-m=1$ and $k+m=n^2$
  • $k-m=n$ and $k+m=n$
  • $k-m=n^2$ and $k+m=1$

Suppose the first bullet is true. Then $m=\frac{(n-1)(n+1)}{2}$, an even number,since $n-1$ and $n+1$ are even. Contradict the fact $m$ is odd. Similarly we get contradictions of latter two. Hence the statement is true.

Is this correct? If not,what I'm doing wrong ?

Edit:I realize my mistake. If $n$ is prime, then my count is correct. Kindly add other information about this to your answer if you wish

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No, that is not correct. We don't know much about $n$, so it is very likely to be composite - which then means that $n^2$ would have more factors than those listed. Factor $n^2$ in some other way, and it all breaks down.

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  • $\begingroup$ Oh sorry,if $n$ is prime, then my count is correct! Am i right? $\endgroup$ – Chinnapparaj R Dec 24 '18 at 11:55
  • $\begingroup$ m and n together with k must make a Pythagorean triples in which one number in left hand side is even and other one is odd. $\endgroup$ – sirous Dec 24 '18 at 12:14
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Every odd positive interger $n$ can be written as $n =2m+1$ where $m$ is an interger and $m \geqslant 0$. So let $\tilde{n}=2n+1$ and $\tilde{m} = 2m+1$. So $ \tilde{m}^2+ \tilde{n}^2 = (2n+1)^2 + (2m+1)^2= 4n^2 +4n +1 +4m^2 + 4m + 1$ =$$2(2n^2+2m^2+2m+2n+1)$$

Call what is inside the parenthesis $\alpha$, so $\sqrt{\tilde{m}^2+ \tilde{n}^2}= \sqrt{ 2\alpha} = \sqrt {2} \sqrt{\alpha}$. Clearly, $\alpha$ id odd, so we can't cancel out the $\sqrt{2}$ factor, so it cant't be a perfect square.

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