# Problem 4. (Kuyanov 9.)

Find all quadruples of natural numbers $a, b, c, d$ for which the following equations are satisfied:

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

Solution:

$\left\{\begin{array}{l}0=c d-a-b \\ 0=a b-c-d\end{array} \Rightarrow(a-1)(b-1)+(c-1)(d-1)=(c d-a-b)+(a b-c-d)+2=2\right.$

Each of $(a-1)(b-1)$ and $(c-1)(d-1)$ is a non-negative integer. If this is:

1 and $1 \Rightarrow(a, b, c, d)=(2,2,2,2)$

0 and $2 \Rightarrow(c-1)(d-1)=2 \Rightarrow\{c, d\}=\{2,3\},\{a, b\}=\{1, x\}, a+b=c d=6 \Rightarrow b=5$

2 and 0 will give a permutation. It remains to check that all the obtained sets satisfy the system of equations.

Criteria: Correct solution: 15 points (+)

No penalty for losing some cases with permutations of $1,2,3,5$

We consider statements of the form $a+b>3$ to be obvious

No verification of the answer (and one-sided consequences in the solution): 13 points (+.)

Reduced to a manageable number of cases (e.g., proven that all numbers are less than 6): 8 points ( + ( $/ 2$ )

Obtained the equality $(a-1)(b-1)+(c-1)(d-1)=(c d-a-b)+(a b-c-d)+2=2$ or similar: 8 points $(+/ 2)$

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