1. Consider the sets $A=\left\{x_{n} \mid x_{n}=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n+57, n \in \mathbb{N}^{*}\right\}$ and $B=\left\{y^{4} \mid y \in \mathbb{N}\right\}$.

a) Prove that $x_{9} \notin B$.

b) Determine the set $A \cap B$.
1. a) The last digit of the number $x_{9}$ is 7, so $x_{9}$ is not a perfect square. Since $B$ contains only perfect squares, it means that $x_{9} \notin B$.

b) If $n \geq 5$, then $1 \cdot 2 \cdot 3 \cdot \ldots \cdot n + 57 = \overline{7} \neq y^{4}, \forall y \in \mathbb{N}$. For $n \in \{1,2,3,4\}$, $x_{n}$ has, respectively, the values 58, 59, 63, and 81. The only one of these numbers that belongs to $B$ is 81, therefore $A \cap B = \{81\}$.