Find $c+d$ when $a,b,c,d\in\mathbb N$, $ab=2(c+d)$, $cd=2(a+b)$ Question:
\begin{align}
&a,b,c,d\in\mathbb N\\
&ab=2(c+d)\\
&cd=2(a+b)\\
&a+b\ge c+d\\
&\text{(Four numbers don't need to be all different.)}\\\\
&\text{Find all possible values for }c+d
\end{align}
What I tried was
\begin{align}
&x^2-(a+b)x+2(c+d)=0\\
&x^2-(c+d)x+2(a+b)=0\\
&(a+b)^2\ge8(c+d)\\
&(c+d)^2\ge8(a+b)\\
&\rightarrow a+b\ge8,\space c+d\ge8\\
\end{align}
which didn't help.

Hint
$$0=ab+cd-2(c+d)-2(a+b)\\
8=(a-2)(b-2)+(c-2)(d-2)\\
$$
This leads to only few possibilities to check where all numbers are at least $2$.
Hint 2: If $a=1$ (or any other of the numbers) you have
$$b=2(c+d)\\
cd=2(b+1)$$
hence
$$cd=2(b+1)=2(2(c+d)+1)=4c+4d+2\\
(c-4)(d-4)=18$$
which is easy to solve by factoriations.
The case where one of the numbers is 0 is trivial.
