The number of solutions to $x^2-xy+y^2=n$ is finite and a multiple of 6.

Any solution $(x,y)$ can be transformed into a different solution $(x-y, x)$. Applying this transform repeatedly gives: $$\begin{align} (x &,y) \\ (x-y &, x)\\ (-y &, x-y) \\ (-x &, -y) \\ (y-x &, -x) \\ (y &, y-x)\\ \end{align}$$

and back to the original - thus the divisibility by $6$.

Clearly the solutions are finite; for example $(x-y)^2 = x^2-2xy+y^2\ge 0$ and thus as $x^2-xy+y^2 > xy$ and since the same-sign case is smaller than the mixed-sign case, we need $|xy|\le n$


Let's multiply the equation by $2$ to get $$ 2x^2-2xy+2y^2=x^2+y^2+(x-y)^2=2n. $$ Clearly $x^2$ and $y^2$ are bounded by $2n$, thus the number of solutions is finite.

Consider the triple $(x,y,z=x-y)$. If all three numbers are different then we can get 6 different solutions by doing all permutations of those and possibly adjusting one sign to negative to satisfy the relation "the last is the first minus the second". Besides those we can get another 6 solutions by changing all the signes. It gives 12 solutions in this group. A special case is when not all three are different, but in this case the number of unique solutions reduces by factor two, and we get a group of only 6 solutions.

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Number Theory

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