
03.2. Find all triples of integers $(x, y, z)$ satisfying

$$
x^{3}+y^{3}+z^{3}-3 x y z=2003
$$


Solution. It is a well-known fact (which can be rediscovered e.g. by noticing that the left hand side is a polynomial in $x$ having $-(y+z)$ as a zero) that

$$
\begin{aligned}
& x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
&=(x+y+z) \frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{2}
\end{aligned}
$$

The second factor in the right hand side is non-negative. It is not hard to see that 2003 is a prime. So the solutions of the equation either satisfy

$$
\left\{\begin{aligned}
x+y+z & =1 \\
(x-y)^{2}+(y-z)^{2}+(z-x)^{2} & =4006
\end{aligned}\right.
$$

or

$$
\left\{\begin{aligned}
x+y+z & =2003 \\
(x-y)^{2}+(y-z)^{2}+(z-x)^{2} & =2
\end{aligned}\right.
$$

Square numbers are $\equiv 0$ or $\equiv 1 \bmod 3$. So in the first case, exactly two of the squares $(x-y)^{2},(y-z)^{2}$, and $(z-x)^{2}$ are multiples of 3. Clearly this is not possible. So we must have $x+y+z=2003$ and $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2$. This is possible if and only if one of the squares is 0 and two are 1's. So two of $x, y, z$ have to be equal and the third must differ by 1 of these. This means that two of the numbers have to be 668 and one 667. A substitution to the original equation shows that this necessary condition is also sufficient.
