IMO Team Selection Test 3

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 $(2,1,1,0)$.