Number of solution for $xy +yz + zx = N$

Question

Is there a way to find number of "different" solutions to the equation $xy +yz + zx = N$, given the value of $N$.

Note: $x,y,z$ can have only non-negative values.

Answer 1

The problem is difficult, as it is related to the determination of class numbers of quadratic number fields. See the references I have given in the comments.

Answer 2

equation:

$XY+XZ+YZ=N$

Solutions in integers can be written by expanding the number of factorization: $N=ab$

And using solutions of Pell's equation: $p^2-(4k^2+1)s^2=1$

$k$ -some integer which choose on our own.

Solutions can be written:

$X=ap^2+2(ak+b+a)ps+(2(a-2b)k+2b+a)s^2$

$Y=2(ak-b)ps+2(2ak^2+(a+2b)k+b)s^2$

$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$

And more:

$X=-2bp^2+2(k(4b+a)+b)ps-2((4b+2a)k^2+(2b-a)k)s^2$

$Y=-(2b+a)p^2+2(k(4b+a)-b-a)ps-(8bk^2-(4b+2a)k+a)s^2$

$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$