10.3. Solve the system of equations in natural numbers

$$
\left\{\begin{array}{l}
a b=c+d \\
c d=a+b
\end{array}\right.
$$
Answer: (1;5;2;3), (1;5;3;2), (5;1;2;3), (5;1;3;2), (2;2;2;2), (2;3;1;5), (2;3;5;1), (3;2;1;5), $(3 ; 2 ; 5 ; 1)$.

Solution: By adding the equations, moving everything to one side, and adding 2 to both sides of the equation, we get

$$
\begin{aligned}
& a b-a-b+1+c d-c-d+1=2 \\
& (a-1)(b-1)+(c-1)(d-1)=2
\end{aligned}
$$

Both addends are non-negative because the factors in them are non-negative.

Therefore, there are 3 cases:

1) $(a-1)(b-1)=0,(c-1)(d-1)=2$. Then one of the numbers $c-1$ and $d-1$ equals 1, and the other equals 2, i.e., one of the numbers $c$ and $d$ equals 2, and the other equals 3. $a+b=c d=6$. From $(a-1)(b-1)=0$ it is clear that $a-1$ or $b-1$ equals 0, i.e., $a=1$ or $b=1$. But then $b=6-1=5$ or $a=6-1=5$. All quadruples $(1 ; 5 ; 2 ; 3),(5 ; 1 ; 2 ; 3),(1 ; 5 ; 3 ; 2),(5 ; 1 ; 3 ; 2)$ of numbers $(a ; b ; c ; d)$ fit.
2) $(a-1)(b-1)=(c-1)(d-1)=1$. Then $a-1=b-1=c-1=d-1=1, a=$ $b=c=d=2$.
3) $(a-1)(b-1)=2,(c-1)(d-1)=0$. This is analyzed similarly to case 1, resulting in the quadruples $(2 ; 3 ; 1 ; 5),(2 ; 3 ; 5 ; 1),(3 ; 2 ; 1 ; 5),(3 ; 2 ; 5 ; 1)$.

Criteria: Only the correct answers 1,2,3,5 and 2,2,2,2 - 1 point. The problem is solved correctly but not all 9 answers are found - 6 points.