Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$

Do you know if there are other integer solutions to
  $$ a+b=c \cdot d \quad \text{ and } \quad  a \cdot b=c+d$$
  besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?


Here's a solution when $a$, $b$, $c$, and $d$ are positive; I'm not sure how to tackle the more general problem though (although something tells me it's probably similar).
Note that substituting $b=cd-a$ into the second equation yields $$a(cd-a)=c+d\implies a^2-cad+c+d=0.$$ By the Quadratic Formula on $a$, we have $$a=\dfrac{cd\pm\sqrt{(cd)^2-4(c+d)}}2.$$ This can only be an integer when $(cd)^2-4(c+d)$ is a perfect square.
Note that since $(cd)^2-4(c+d)\equiv (cd)^2\pmod 2$, we have $$(cd)^2-4(c+d)\leq(cd-2)^2.$$  This rearranges to $$-4(c+d)\leq -4cd+4\implies 2\geq (c-1)(d-1).$$  This is great news, because it means we actually don't have many cases to check!
CASE 1: $c=1$ (or $d=1$, for that matter).
Plugging back in, we see that for some integer $K$ we have $$d^2-4(d+1)=(d-2)^2-8=K^2.$$  The only case that works here is $d=5$ (since $d=6$ fails and past that point perfect squares differ by at least $9$).  By symmetry, $c=5$, $d=1$ also works.
CASE 2: $c=2$ and $d\leq 3$.
Plugging back in to the original expression, we need $$(2d)^2-4(2+d)=(2d-1)^2-9$$ to be a perfect square.  Trial and error works since $d$ is small, but here $d=2$ and $d=3$ both work.
CASE 3: $c=3$ and $d=2$.
Oh look, we've already shown this works!  (We don't need to check $d=1$ because that was already covered in Case 1.)
Hence the only possible pairs for $(c,d)$ are $(2,3)$, $(2,2)$, $(5,1)$, arrangements. By a similar argument, we see that the same exact possible pairs are the only possible ones for $(a,b)$ as well.
From here, it's not hard to see that $(a,b,c,d)=\boxed{(2,2,2,2),(2,3,5,1)}$ and permutations are the only possible solutions for $a,b,c,d>0$.
