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:

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

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

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

Answer 3

Perhaps these formulas for someone too complicated. Then equation:

$XY+XZ+YZ=N$

If we ask what ever number: $p$

That the following sum can always be factored: $p^2+N=ks$

Solutions can be written.

$X=p$

$Y=s-p$

$Z=k-p$

Answer 4

the equation: $XY+XZ+YZ=a(X+Y+Z)$

Solutions can be written using the solutions of Pell's equation:

$p^2-(k^2-k+1)s^2=a$

$k$ - This number can be any and defined by us.

Then the decision will be Met form:

$X=p^2+(k+1)ps$

$Y=p^2+(k-2)ps$

$Z=p^2+(1-2k)ps$

And more.

$X=(k+1)ps-(k^2-k+1)s^2$

$Y=(k-2)ps-(k^2-k+1)s^2$

$Z=(1-2k)ps-(k^2-k+1)s^2$

Answer 5

Here is a C# program that does the job in a pretty naive way. It eliminates some possibilities, then bruteforces all solutions. You can also use the Mathematica code provided on the OEIS page on this sequence (thank Gerry Myerson for this!).

using System;

class Program
{
    static void Main(string[] args)
    {
        Console.Write("N = ");
        int N = int.Parse(Console.ReadLine());
        Console.WriteLine();

        int u = 0, add = -1, n = 0, s = 0;
        while (u < 2 * N) { u += (add += 2); n++; }
        while (n <= N + 1)
        {
            if (canBeWrittenAsSumOfThreeSquares(u - 2 * N))
            {
                for (int x = 0; x <= n / 3; x++)
                    for (int y = x; y <= (n - x) / 2; y++)
                    {
                        int z = n - x - y;
                        if (x * y + y * z + z * x == N)
                        {
                            Console.Write(x + ", " + y + ", " + z + " ");
                            if (x != y && y != z && x != z) { Console.WriteLine("(6 permutations)"); s +=     6; }
                            else if (x == y && y == z) { Console.WriteLine("(1 permutation)"); s++; }
                            else { Console.WriteLine("(3 permutations)"); s += 3; }
                        }
                    }
            }
            u += (add += 2);
            n++;
        }
        Console.WriteLine(s + " solutions found.");
        Console.ReadKey(true);
    }

    static bool canBeWrittenAsSumOfThreeSquares(int n)
    {
        while (n % 4 == 0 && n > 0) n /= 4;
        return n % 8 != 7;
    }
}

The code uses the fact that $xy + yz + zx = \frac{(x + y + z)^2 - x^2 - y^2 - z^2}{2}$. When we have $xy + yz + zx = N$, we have $(x + y + z)^2 - x^2 - y^2 - z^2 = 2N$ and $ (x + y + z)^2 - 2N = x^2 + y^2 + z^2 $ a number can be written as a sum of three squares iff it is not of the form $4^k(8m + 7)$ see wikipedia article. So we simply check all $u \in \mathbb{N}$ so that $0 \leq u^2 - 2N \leq N^2 $, and then iterate through all increasing ($x \leq y \leq z$) 3-tuples $(x, y, z)$, multyplying with the number of permutations that is possible (either 1, if all three components are the same, 3, when two components are the same and one different, or 6, when all the components are different).

I haven't used any advanced theorem (class numbers?), but I just wanted to illustrate that there is an approach that works for small $N$. I was able to calculate the amount of solutions for $N=4000$ in about one minute. This approach probably can be expanded to be more efficient.