2023/24 AHSMC Part 1 Problem 5

Observe that $x = 2$ satisfies the inequality. When $x \ne 2$ then $x$ must satisfy $$\frac{(x-2)^2(x+2)^2}{(2-x)^2} \le 1,$$ that is $(x+2)^2 \le 1$. Thus $-3 \le x \le -1$. There are four integer solutions: $x = -3,-2,-1,2$. The answer is (E).