10. How many quadruples of positive integers $(a, b, x, y)$ are there such that $x+y=a \cdot b$ and $a+b=x \cdot y$?
(A) 1
(B) 5
(C) 9
(D) 17
(E) Infinite
10. The answer is (C). If $a, b, x, y$ are all greater than or equal to 2, then $a b \geq a+b=x y \geq x+y=a b$, so each inequality must be an equality and the only possible quadruple is $(2,2,2,2)$. If instead one of the four numbers is equal to 1, say for example $b=1$, then substituting the equality $a=x+y$ into $a+1=x y$ yields $(x-1)(y-1)=2$, from which we get $x=2, y=3$ or vice versa. One solution is thus $(a, b, x, y)=(1,5,2,3)$, and the others are obtained by swapping the roles of $a, b$, the roles of $x, y$, or the roles of the pair $(a, b)$ and the pair $(x, y)$: this yields another eight solutions, namely $(1,5,2,3),(1,5,3,2),(5,1,2,3)$, $(5,1,3,2),(2,3,1,5),(2,3,5,1),(3,2,1,5),(3,2,5,1)$.