Find Integral Solutions of $m^2+(m+1)^2 = n^4+(n+1)^4$

Question

Find integral solutions for $m$ and $n$. $$ m^2 + (m+1)^2 = n^4 + (n+1)^4. $$

It is question from a mathematics magazine. I had already found four solutions (which are trivial) $m = \{-1,0\} .$ Is there any other Integer. If not how can I prove it.

Answer 1

If $n$ is negative then we can replace $n$ by the positive integer $-1-n$. So suppose $n$ is non-negative. Similarly we can suppose $m$ is non-negative.

Now consider $m=n^2+n$ and $m=n^2+n+1$. The respective values of $m^2+(m+1)^2$ are less than and greater than $n^4+(n+1)^4$ except in the one case that $n=0$.

So all solutions have $n=0$ or $-1-n=0$ i.e. $n\in \{0,-1\}$. Then $m\in \{0,-1\}$ also.