## Task B-1.1.

For distinct real numbers $a$ and $b$, the following equations hold:

$$
a+9=(b-3)^{2} \quad \text { and } \quad b+9=(a-3)^{2}
$$

What is the value of $a^{2}+b^{2}$?
## Solution.

We will square the binomials and simplify the given equations, then add and subtract them. We get

$$
\begin{aligned}
a+9 & =b^{2}-6 b+9 \\
b+9 & =a^{2}-6 a+9
\end{aligned}
$$

or equivalently,

$$
\begin{aligned}
& a=b^{2}-6 b \\
& b=a^{2}-6 a
\end{aligned}
$$

After adding, we get

$$
a+b=a^{2}+b^{2}-6(a+b)
$$

or equivalently,

$$
a^{2}+b^{2}=7(a+b)
$$

After subtracting, we get

$$
\begin{aligned}
a-b & =b^{2}-a^{2}-6(b-a) \\
& =-(a-b)(a+b)+6(a-b)
\end{aligned}
$$

from which, by dividing by $a-b \neq 0$, we obtain

$$
1=-(a+b)+6, \quad \text { i.e. } \quad a+b=5
$$

Finally, from (1) we get $a^{2}+b^{2}=7(a+b)=7 \cdot 5=35$.