How many solutions are there to $F(n,m)=n^2+nm+m^2 = Q$?

Question

Let $n,m$ be two positive integers, we consider: $$F(n,m)=n^2+nm+m^2$$

Let $Q$ be one value reach by $F(n,m)$.

How many different pairs $(n,m)$ verify $F(n,m)=Q$?

Answer 1

As it turns out, we can give a complete answer to this question. The exact number of solutions depends on the prime factorization of $Q$ Specifically, it is a function of the exponents of the prime factors which are congruent to $1$ mod $3$, with the condition that all the factors congruent to $2$ modulo $3$ have their prime factors appear with even multiplicity.
Note: I have not yet provided a proof, the result that primes of the form $1+3k$ can be represented is a theorem of Jacobi.

Let $$Q=\prod_i q_i^{\alpha_i} \prod_i p_i^{\beta_i}$$ where the $q_i$ are $1$ mod $3$ and the $p_i$ are $2$ mod $3$. Our equation $n^2+nm+m^2$ has solutions if and only if each $\beta_i$ is even.

Proof: Take the equation modulo $3$. By case analysis for $n,m$ the right hand side cannot be congruent to $2$, and hence the statement follows.

Remark: Notice the following similarity to the sum of squares problem (points on a circle).

The Answer: Suppose that as before $$Q=\prod_i q_i^{\alpha_i} \prod_i p_i^{\beta_i}$$ where the $q_i$ are $1$ mod $3$ and the $p_i$ are $2$ mod $3$. Suppose as well that all of the $\beta_i$ are even. (Otherwise we can have no solutions)

Let $$B=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1).$$ Then the number of non-negative integer solutions to $$m^2+mn+n^2=Q$$ is exactly $$\left\lceil\frac{B}{2}\right\rceil.$$
(Again notice the similarity to the sum of squares function)

In particular, if $l(Q)$ is the number of representations where $n,m$ are any integers, (that is positive or negative) then $$l(Q)=2B.$$

Hope that helps.