11.2. Find all natural $n$ that can be represented as the sum $n=a^{2}+b^{2}$, where $a$ is the smallest divisor of $n$ different from 1, and $b$ is some divisor of $n$.
Answer: 8 and 20.

Solution. If $n$ is odd, then all its divisors are odd. Therefore, the right side of the equation $n=a^{2}+b^{2}$ is even - a contradiction. Hence, $n$ is even and its smallest non-unit divisor $a$ is 2, so $n=4+b^{2}$. By the condition, $b$ divides $n=4+b^{2}$, which means it also divides the difference $n-b^{2}=4$, so $b$ must be one of the numbers $1,2,4$. In this case, $n$ equals $5,8,20$ respectively. The first case is not valid due to the oddness, the other two satisfy the problem's condition.

Grading criteria. Proved that $n$ is even: 1 point. Missed one of the solutions: minus 2 points. Only guessed both answers with verification: 1 point.