2015 AMC 10A Problem 20

Let the side lengths be positive integers $x$ and $y$. Then $$\begin{aligned} &A+P=xy+2x+2y \\ &=(x+2)(y+2)-4. \end{aligned}$$ Hence $A+P+4$ must factor into two integers both at least $3$. The answer choices plus $4$ are $104,106,108,110,112$. All except $106$ have a factorization with both factors at least $3$: $$104=4\cdot26,$$ $$108=9\cdot12,$$ $$110=10\cdot11,$$ $$112=7\cdot16.$$ But $106=2\cdot53$, so it cannot equal $(x+2)(y+2)$. Thus, B is the correct answer.