I am considering the following equation $$N(n^2+m^2+nm)-(k^2+l^2+kl)=0.$$ where $N\ge2$ is an integer. I would like to find for which $N$ there exist a set of integers $n,m,k,l$ satisfying this equation (I know there is a solution for instance for $N=1$ and $N=3$).
Numerics and other considerations however suggest there is no such set of integers for $N=2$, but I would like to have a rigorous proof. I have tried playing a bit with modular arithmetic but I didn't have much success. Can anyone come up with a proof for the case $N=2$?
Hopefully the same process can then be applied to higher values of $N$...
For $N=2:$ The exponent of $2$ when factoring $2(n^2+m^2+nm)$ will be odd, while the exponent of $2$ when factoring $(k^2+l^2+kl)$ will be even. They cannot be equal.
Still with the fixed $N=2.$ We do not really need quadratic reciprocity for this. Indeed, $x^2 + xy + y^2$ is odd unless both $x,y$ are even. If both are even, we can divide both by 2, with the result that we have divided $x^2 + xy + y^2$ by $4.$ Do this as many time as necessary until at least one of the variables is now odd. The result is that the exponent of $2$ in prime factorization of the original $x^2 + xy + y^2$ was even!
There is a solution if and only if we can express $$ N = u^2 + uv + v^2 $$ in integers.
This is the same as saying that, whenever a prime $q \equiv 2 \pmod 3$ and $N$ is divisible by $q,$ then the exponent of $q$ in the prime factorization of $N$ is even.
The same characterization applies to both $m^2 + mn + n^2$ and $k^2 + kl + l^2.$
In turn, this is the same as saying that $h(-3) = 1,$ there is only one equivalence class of (positive) binary quadratic forms of discriminant $-3.$