11. (CAN 1) Let $n$ be a natural number and $a_{1}, a_{2}, \ldots, a_{2 n}$ mutually distinct integers. Find all integers $x$ satisfying
$$
\left(x-a_{1}\right) \cdot\left(x-a_{2}\right) \cdots\left(x-a_{2 n}\right)=(-1)^{n}(n!)^{2} .
$$
11. Suppose that an integer $x$ satisfies the equation. Then the numbers $x-$ $a_{1}, x-a_{2}, \ldots, x-a_{2 n}$ are $2 n$ distinct integers whose product is $1 \cdot(-1)$. $2 \cdot(-2) \cdots n \cdot(-n)$.
From here it is obvious that the numbers $x-a_{1}, x-a_{2}, \ldots, x-a_{2 n}$ are some reordering of the numbers $-n,-n+1, \ldots,-1,1, \ldots, n-1, n$. It follows that their sum is 0 , and therefore $x=\left(a_{1}+a_{2}+\cdots+a_{2 n}\right) / 2 n$. This is
the only solution if $\left\{a_{1}, a_{2}, \ldots, a_{2 n}\right\}=\{x-n, \ldots, x-1, x+1, \ldots, x+n\}$ for some $x \in \mathbb{N}$. Otherwise there is no solution.