## Task A-3.1.

Determine all quadruples $(a, b, c, d)$ of natural numbers for which

$$
a+b=c d, \quad a b=c+d
$$
## Solution.

If the values $c d=a+b$ and $a b=c+d$ are equal, then $a b=a+b$ implies $(a-1)(b-1)=1$. Since $a$ and $b$ are natural numbers, it must be that $a=b=2$. Similarly, $c=d=2$, so one solution is $(a, b, c, d)=(2,2,2,2)$.

If the values $c d=a+b$ and $a b=c+d$ are not equal, without loss of generality, assume that $a b < c d$ (i.e., $a b < a+b$).

Thus, we obtain all solutions:

$$
\begin{aligned}
(a, b, c, d)= & (2,2,2,2) \\
& (1,5,2,3),(1,5,3,2) \\
& (5,1,2,3),(5,1,3,2) \\
& (2,3,1,5),(3,2,1,5) \\
& (2,3,5,1),(3,2,5,1)
\end{aligned}
$$