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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 $N$...

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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 $-3.$

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  • $\begingroup$ @LucaMingarelli have you taken any number theory courses? Are you familiar with quadratic reciprocity? $\endgroup$ – Will Jagy Jul 9 '17 at 0:48
  • $\begingroup$ No, to be fair I have never taken number theory and never heard of quadratic reciprocity but I am trying to read something about it from the moment you mentioned it. Sorry I feel a bit lost... What do you mean by $h(-3)$? $\endgroup$ – semola Jul 9 '17 at 0:51
  • $\begingroup$ could you maybe help me to see why this is true for the case $N=2$? $\endgroup$ – semola Jul 9 '17 at 0:57
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    $\begingroup$ @LucaMingarelli in that case I recommend an older book (1929), Introduction to the Theory of Numbers , by Leonard Eugene Dickson. Notice that, using the equivalent formulation i gave about even exponents, the product of two numbers that each has an expression of type $x^2 + xy + y^2$ is a third number that also can be expressed in that manner. That is your original question, put briefly $\endgroup$ – Will Jagy Jul 9 '17 at 0:59
  • $\begingroup$ thank you very much for your answer. Can I ask you one last thing? Are you aware of a similar result for the case in which $N$ is allowed to be real rather then integer? $\endgroup$ – semola Jul 9 '17 at 15:31

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