$\forall m\in\mathbb N\exists n>m+1\exists N\in\mathbb N:m^2+n^2+(mn)^2=N^2$ Prove the conjecture or give a counter-example:
For each $m\in\mathbb N$ there exist a $n>m+1$ such that $m^2+n^2+(mn)^2$ is a perfect square.
I have just tried it out numerically and it holds for $m<1000$.
I can't see any pattern for the smallest $n$:
m  n

1 12
2 8
3 18
4 32
5 50
6 72
7 98
8 30
9 162
10 200
11 242
12 119
13 338
14 392
15 450
16 512
17 578
18 105
19 722
20 800
21 208
22 968
23 1058
24 1152
25 1250


I might as well turn my comments into an answer...
For $m>1$ it is easy to see $n=2m^2$ works. Indeed, $n>m^2>m+1$ and
$$m^2+n^2+(mn)^2=m^2+4m^4+4m^3=m^2(1+4m+4m^2)=m^2(1+2m)^2=(m+2m^2)^2.$$
