$\forall m\in\mathbb N\exists n>m+1\exists N\in\mathbb N:m^2+n^2+(mn)^2=N^2$

Question

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

Answer 1

Your conjecture is true.

Fix $m\ge 1$. You need that $n^2(m^2+1)+m^2$ is a square. It would be enough to find such a $n$ with the additional condition that it is divisible by $m$, let us say $n=mA$. After dividing by $m^2$, it would be enough to find positive integers $A,B$ (with $A$ sufficiently large) such that $$ (m^2+1)A^2-B^2=1. $$ Start with the basic solution $(1,m)$, and then construct infinitely many ones with Pell's equation method (see, e.g., equation (37) here).

Answer 2

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 $$m^2+n^2+(mn)^2=m^2+4m^4+4m^3=m^2(1+4m+4m^2)=m^2(1+2m)^2=(m+2m^2)^2.$$