10) For how many ordered pairs $(a, b)$ of integers does their product equal their sum?
(A) None,
(B) one,
(C) two,
(D) four
(E) more than four.
10. The answer is $\mathbf{( C )}$. First solution. The equality $a+b=ab$ is equivalent to $(a-1)(b-1)=1$, so the factors $a-1$ and $b-1$ must both be 1 or both be -1. The first case leads to the pair $(a=2, b=2)$; the second case leads to the pair $(a=0, b=0)$. The pairs that satisfy the required property are a total of two.

Second solution. We must have $ab=a+b$; observe that the pair $(0,0)$ satisfies this equality. Furthermore, if $a=0$ then $0=0+b$ and thus $b=0$; similarly, if $b=0$ then $a=0$. Therefore, if we want to find other solutions besides $(0,0)$, we can assume that both $a$ and $b$ are non-zero. Then $b=1+\frac{b}{a}$, so $a$ divides $b$; similarly, $a=\frac{a}{b}+1$ and thus $b$ divides $a$. The only possibility is then that $a= \pm b$. If $a=b$ then $a^2=2a$ which means $a=2$ and we get a second solution given by $(2,2)$. If $a=-b$ then $-a^2=0$ which means $a=0$, but this brings us back to the solution $(0,0)$ already found. In conclusion, the pairs that satisfy the required property are two: $(0,0)$ and $(2,2)$.