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How do we prove that $$n^2+m^2 = a^2+b^2 = (n-a)^2+(m-b)^2$$ has no nonzero integer solutions? I know two ways to prove this by taking a geometric interpretation but I don't want such a version. How can I prove the above by just "looking at the equations"? What is the "natural" approach.

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    $\begingroup$ Related: math.stackexchange.com/questions/105330/…. Note that the accepted solution is purely algebraic. $\endgroup$ – Micah Jun 20 '14 at 16:50
  • $\begingroup$ @Micah, that's exactly the motivation behind this problem but I saw it elsewhere. One can use Pick's theorem or draw the smallest enclosing rectangle and either way argue to contradiction that all areas must be rational. $\endgroup$ – abnry Jun 20 '14 at 16:58
  • $\begingroup$ Hey can you please explain the geometrical interpretation approach to solve this problem? $\endgroup$ – user3628041 Jun 23 '14 at 9:40
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Recall that $p^2+q^2$ is of the form $4k$ if $p,q$ are both even, of the form $4k+1$ if they are of different parities, and of the form $4k+2$ if they are both odd.

Let $(n,m,a,b)$ be a solution where $n > 0$ is minimized. As $$n^2+m^2=(n^2+m^2)+(a^2+b^2)-[(n-a)^2+(m-b)^2]=2an+2bm \, ,$$ $n^2+m^2$ is even.

  • If $n^2+m^2$ is a multiple of $4$, then $n,m,a,b$ are all even, so we can obtain a solution with smaller $n$ by dividing all of them by $2$: a contradiction.
  • If $n^2+m^2$ is of the form $4k+2$, then $n,m,a,b$ are all odd. But then $n-a$ and $m-b$ are even, and so $(n-a)^2+(m-b)^2$ is a multiple of $4$: also a contradiction.

So there is no solution with $n>0$.

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From individ's comment, $n^2+m^2$ is even, so $n$ and $m$ are both even or both odd. So are $a$ and $b$.

If $a,b,m,n$ are all odd, then $2an+2mb$ is a multiple of 4, but $n^2+m^2$ is not.

If $n$ and $m$ are even, then $2an+2bm$ is a multiple of 4, so $a^2+b^2$ is a multiple of 4, so $a$ and $b$ are even.

Then $a,b,m,n$ are all even, so have a common factor, 2.

Divide all of $a,b,m,n$ by 2, and you get a smaller solution in integers.

So there is no smallest solution. Hence no non-zero solution.

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If this is done, it will run the next system.

$$n^2+m^2=a^2+b^2=2an+2bm$$

And it can't be.

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  • $\begingroup$ I know you'll get this system, but I don't know why this cannot have a nonzero solution. Maybe it is simple but I haven't thought about it. $\endgroup$ – abnry Jun 20 '14 at 16:20

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