Solve in positive integers the equation $x^3 + y^3 + z^3 - 3xyz = 1517$

Question

Solve in positive integers the equation

$$x^3 +y^3 +z^3 −3xyz = 1517$$

Initially, I tried to factorize the LHS and my last step was

$$(x+y+z)(x^2+ y^2 + z^2 -xy -zx - yz) = 37×41$$

How do I find the value of $x, y, z$? Any help would be appreciated

Answer 1

To continue your argument, we have the following cases:

1. $x + y + z = 1, x^2 + y^2 + z^2 - xy - yz - zx = 1517$;
2. $x + y + z = 37, x^2 + y^2 + z^2 - xy - yz - zx = 41$;
3. $x + y + z = 41, x^2 + y^2 + z^2 - xy - yz - zx = 37$;
4. $x + y + z = 1517, x^2 + y^2 + z^2 - xy - yz - zx = 1$.

Case 1 is impossible, as $x + y + z \geq 3$.

The three remaining cases can be done in the same way. I will do case 3 as example.

We rewrite the second equality as $(x - y)^2 + (y - z)^2 + (z - x)^2 = 74$. There are only three ways to write $74$ as a sum of three squares: $$74 = 0^2 + 5^2 + 7^2 = 1^2 + 3^2 + 8^2 = 3^2 + 4^2 + 7^2.$$ Since the sum $(x - y) + (y - z) + (z - x)$ must be zero, we see that the only possibilities, up to permutation of $x, y, z$, are $$x - y = 3, y - z = 4, z - x = -7$$ or $$x - y = 3, y - z = -7, z - x = 4.$$ Together with $x + y + z = 41$, we get $(x, y, z) = (17, 14, 10)$.

With similar calculations, we see that there is no solution in case 2, and $(x, y, z) = (506, 506, 505)$ in case 4.

Therefore all solutions are given by $(x, y, z) = (17, 14, 10), (506, 506, 505)$ up to permutation.