How to show $n^2+m^2 = a^2+b^2 = (n-a)^2+(m-b)^2$ has no nonzero integer solutions? 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.
 A: 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 $n>0$.
A: 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.
A: 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.
