## Task 1.

Determine all pairs $(a, b)$ of real numbers for which

$$
a+\frac{a+8 b}{a^{2}+b^{2}}=2 \quad \text { and } \quad b+\frac{8 a-b}{a^{2}+b^{2}}=0
$$
## Solution.

Notice that $a$ and $b$ cannot both be zero at the same time, as this would make the denominator $a^{2}+b^{2}$ also zero.

If $a=0$, then from the first equation we get $b=4$, and from the second equation we get $b=1$ or $b=-1$, which cannot be true simultaneously. Similarly, if $b=0$, then from the second equation we get $\frac{8 a}{a^{2}}=0$, which is also impossible.

By multiplying the first equation by $b$ and the second by $a$ and then adding them, we get

$$
a b+\frac{a b+8 b^{2}}{a^{2}+b^{2}}+a b+\frac{8 a^{2}-a b}{a^{2}+b^{2}}=2 b
$$

which simplifies to $2 a b+8=2 b$.

By multiplying the first equation by $a$ and the second by $b$ and then subtracting the second from the first, we get

$$
a^{2}+\frac{a^{2}+8 a b}{a^{2}+b^{2}}-b^{2}-\frac{8 a b-b^{2}}{a^{2}+b^{2}}=2 a
$$

which simplifies to $b^{2}=a^{2}-2 a+1=(a-1)^{2}$. We consider two cases.

In the first case, when $b=1-a$, substituting into $2 a b+8=2 b$ gives

$$
2(1-b) b+8=2 b
$$

from which we get $b^{2}=4$. Therefore, in this case, $b=2$ and $a=-1$ or $b=-2$ and $a=3$.

In the second case, when $b=a-1$, we similarly get $b^{2}=-2$, which has no solutions in real numbers.

Finally, the solutions are $(a, b) \in\{(-1,2),(3,-2)\}$.