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

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

Suppose that $n=2k+1$ and $m=2l+1$ for some integers $k$ and $l$. Suppose that $m^2+n^2$ is a perfect square. Then there exists an integer $t$ such that $t^2=m^2+n^2=4l^2+4l+1+4k^2+4k+1=2(2l^2+2k^2+2k+2l+1)$. Then $t$ is even or $t=2r$. $4r^2=2(2l^2+2k^2+2k+2l+1)$ or $2r^2=2l^2+2k^2+2k+2l+1$. RHS is odd but LHS is even. It is a contradiction here. Therefore, $m^2+n^2$ is not a perfect square.