A geometry problem by arko roychoudhury

Let the positive integer side lengths of the rectangle be $x,y$ . Then $A = xy$ and $P = 2x + 2y$ , and so

$A + P = xy + 2x + 2y = (x + 2)(y + 2) - 4 \Longrightarrow A + P + 4 = (x + 2)(y + 2)$ .

So since $x,y \gt 0$ , we must be able to represent $A + P + 4$ as a product $m \times n$ with integers $m,n \gt 2$ . Now as examples for the given options we have that

$108 + 4 = 112 = 8 \times 14, 104 + 4 = 108 = 9 \times 12, 100 + 4 = 104 = 8 \times 13$ and $106 + 4 = 10 \times 11$ , and so it is possible for $A + P$ to be any of these options.

However, the only possibility for $102 + 4 = 106$ is $2 \times 53$ , as both $2$ and $53$ are prime, which does not satisfy the requirement that $m,n \gt 2$ . Thus $A + P$ cannot equal $\boxed{102}$ .