Problem 11.4. Find the least positive integer $a$ such that the system

$$
\left\lvert\, \begin{aligned}
& x+y+z=a \\
& x^{3}+y^{3}+z^{2}=a
\end{aligned}\right.
$$

has no an integer solution.

Oleg Mushkarov
11.4. When $a=1,2,3$ the system has solutions $(1,0,0),(1,1,0)$ and $(1,1,1)$, respectively. We shall prove that when $a=4$ the system has no integer solution.

Suppose the contrary. Then we have

$$
4-z^{2}=x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=(4-z)\left(x^{2}-x y+y^{2}\right)
$$

giving (since $z=4$ does not lead to an integral solution) that $\frac{4-z^{2}}{4-z}$ is an integer. Since $\frac{4-z^{2}}{4-z}=\frac{-12+16-z^{2}}{4-z}=4+z+\frac{12}{z-4}$, we conclude that $z-4$ is a divisor of 12 . Hence $z-4= \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ and therefore $z=-8,-2,0,1,2,3,5,6,7,8,10$ or 16 . Using (1) we have

$$
(x+y)^{2}-3 x y=\frac{4-z^{2}}{4-z} \Longleftrightarrow 3 x y=(4-z)^{2}-\frac{4-z^{2}}{4-z}
$$

and we obtain the following system for $x$ and $y$ :

$$
\left\lvert\, \begin{aligned}
& x+y=4-z \\
& x y=\frac{(4-z)^{3}+z^{2}-4}{3(4-z)}
\end{aligned}\right.
$$

It is easy to check that all the values of $z$ listed above do not lead to an integral solution for $x$ and $y$.