Problem 702

B3. 48 and 60 The number of squares in the length of the rectangle we call $a$ and the number in the width we call $b$. We may assume that $a \geqslant b$. The total number of squares in the rectangle is $a b$ and the number of squares on the edge is equal to $2 a+2 b-4$. Given that half of the squares are on the edge, so $a b=2(2 a+2 b-4)$. Rewriting gives: $a b-4 a-4 b+16=8$. The left side can be factored, so we find that $(a-4)(b-4)=8$.

Since $a$ and $b$ are positive integers and $a \geqslant b$, the only possibilities are: $a-4=$ $8, b-4=1$ and $a-4=4, b-4=2$. The possibilities where $a-4$ and $b-4$ are negative are excluded, because then $b-4$ is at most -4 and $b$ is not positive. In other words: $a=12, b=5$ and $a=8, b=6$. For the rectangle, this gives $12 \times 5=60$ or $8 \times 6=48$ squares in total.