Berkeley Math Circle Monthly Contest 6

Solution:

This is clear for $n = 2$. We use the relation
$$\begin{gathered} a_{n-1} = a_{1} a_{2} \cdots a_{n-2} + 4 \\ a_{1} a_{2} \cdots a_{n-2} = a_{n-1} - 4 \end{gathered}$$
for $n \geq 3$ to transform $a_{n}$:
$$\begin{aligned} a_{n} & = a_{1} a_{2} \cdots a_{n-2} a_{n-1} + 4 \\ & = \left(a_{n-1} - 4\right) a_{n-1} + 4 \\ & = a_{n-1}^{2} - 4 a_{n-1} + 4 \\ & = \left(a_{n-1} - 2\right)^{2}. \end{aligned}$$
This is clearly the square of an integer.