## Task A-3.2.

Determine all pairs of natural numbers $(a, b)$ such that

$$
\begin{gathered}
a^{3}-3 b=15 \\
b^{2}-a=13
\end{gathered}
$$

Note: The only solution to the given system is $(a, b)=(3,4)$. If a student writes that the pair $(a, b)=(3,4)$ is a solution to the given system but does not write any other conclusion that earns points, they should receive a total of 1 point.
## First Solution.

If we add the equations, we get

$$
a^{3}-a+b^{2}-3 b=28
$$

For $a \geqslant 4$, we have $a^{3}-a \geqslant 64$, and $b^{2}-3 b \geqslant-2$ for all natural numbers $b$, so it must be that $a \leqslant 3$.

5 points

1 point

If $a=2$, we get the quadratic equation $b^{2}-3 b=22$ which has no solutions in natural numbers.

If $a=3$, we get the quadratic equation $b^{2}-3 b=4$, which has one natural solution $b=4$ (the other is $b=-1$).

Thus, the pair $(a, b)=(3,4)$ is a solution to the equation obtained by adding the equations of the given system.

Since $3^{3}-3 \cdot 4=15$ and $4^{2}-3=13$, we see that the pair $(a, b)=(3,4)$ is a solution to the given system.

Note: Alternatively, from $a^{3}-a+b^{2}-3 b=28$, we can conclude that $b \leqslant 4$. This conclusion is worth 4 points (while the conclusion $a \leqslant 3$ is worth 5 points). Each of the cases $b=1, b=2$, and $b=3$ is worth 1 point. The conclusion that with $b=4$ it must be $a=3$ is worth 1 point, and verifying that the pair $(a, b)=(3,4)$ satisfies both equations of the system is also worth 1 point.