<s>[INST] Solve in positive integers the equation $x^3 + y^3 + z^3 - 3xyz = 1517$ 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
 [/INST] To continue your argument, we have the following cases:

*

*$x + y + z = 1, x^2 + y^2 + z^2 - xy - yz - zx = 1517$;

*$x + y + z = 37, x^2 + y^2 + z^2 - xy - yz - zx = 41$;

*$x + y + z = 41, x^2 + y^2 + z^2 - xy - yz - zx = 37$;

*$x + y + z = 1517, x^2 + y^2 + z^2 - xy - yz - zx = 1$.

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.
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