Problem 2119

Note that $a > c$ and $b > d$, as $a - c$ and $b - d$ are sides of a non-degenerate triangle. So $a \ge c + 1$ and $b \ge d + 1$, as they are integers. Consider two cases: $a > 2c$ and $a \le 2c$.

Suppose that $a > 2c$. Then $ab > 2bc \ge 2c \cdot (d+1) = 2cd + 2c$. We also have $ab = 2 + 2cd$, so $2c < 2$, and therefore $c = 0$. We deduce that $ab = 2$ and that there exists a non-degenerate triangle with sides $a$, $b-d$, and $d$. Therefore $d \ge 1$ and $b > d$, so $b \ge 2$. From $ab = 2$ it follows that $a = 1$ and $b = 2$, and therefore also $d = 1$. Note that there exists a non-degenerate triangle with sides $1$, $1$, and $1$, so the quadruple $(1, 2, 0, 1)$ is a solution.

Now suppose that $a \le 2c$. By the triangle inequality, we have $(a-c) + (b-d) > c+d$, so $a+b > 2(c+d)$. As $a \le 2c$, it follows that $b > 2d$. As $a \ge c+1$, we have $ab > (c+1) \cdot 2d = 2cd + 2d$. On the other hand, we have $ab = 2+2cd$, so $2d < 2$, and therefore $d = 0$. Analogously to the previous case, we deduce that the only other solution is $(2, 1, 1, 0)$.

Therefore the only solutions are the quadruples $(1, 2, 0, 1)$ and $(2, 1, 1, 0)$.