5. Solution. The second equation can be transformed into the form $(x-3)^{2}+y^{2}=1$. Therefore, if the pair of numbers $\left(3+t_{0} ; y_{0}\right)$ is a solution to the system, then the pair $\left(3-t_{0} ; y_{0}\right)$ will also be a solution. This means that the only possible solutions are $(3 ; 1)$ or $(3 ;-1)$. The first pair is a solution when $a=1$ or $a=4$, and the second pair is a solution when $a=2$ or $a=3$. When $a=1$ or $a=4$, the system reduces to

$$
\left\{\begin{array}{l}
y-1=2(x-3)^{6}+|x-3| \\
(x-3)^{2}+y^{2}=1
\end{array}\right.
$$

and has exactly one solution $(3 ; 1)$, because for $x \neq 3$, the first equation implies $y>1$, which is impossible. When $a=2$ or $a=3$, the system becomes

$$
\left\{\begin{array}{l}
y+1=|x-3| \\
(x-3)^{2}+y^{2}=1
\end{array}\right.
$$

From which we find three solutions: $(3 ;-1),(4 ; 0),(2 ; 0)$.
Answer: 1 and 4 (the answer is the same for option $2-2$).

Answer: $x=\frac{2 \pi k}{7}, x= \pm \frac{\arccos ((\sqrt{3}-1) / 2)}{7}+\frac{2 \pi l}{7}, k, l \in \mathbb{Z}$.

## Lomonosov Moscow State University

 Olympiad "Conquer Sparrow Hills"

Variant 3-1 (Voronezh)