\[
\left[\begin{array}{l}
\text { Equations in integers } \\
{[\text { Case enumeration }}]
\end{array}\right]
\]

Solve the system in natural numbers

\[
\begin{aligned}
& x+y=z t, \\
& z+t=x y .
\end{aligned}
\]
Let's consider two cases.

1) $x=1$. Then $z+t=y=z t-1$, from which $(z-1)(t-1)=z t-z-t+1=2$. Therefore, $\{z, t\}=\{2,3\}, y=6-$ $1=5$.
2) $x, y, z, t>1$. Then $(x-1)(y-1) \geq 1$, that is, $x y \geq x+y$. Similarly, $z t \geq z+t$. Adding these, we get $x y+z t$ $\geq x+y+z+t$. On the other hand, adding the equations of the given system, we get $x y+z t=x+y+z+t$.

Therefore, both inequalities $x y \geq x+y$ and $z t \geq z+t$ are actually equalities, that is, $x=y=z=t=2$.

## Answer

$(1,5,2,3),(5,1,2,3),(1,5,3,2),(5,1,3,2),(2,3,1,5),(2,3,5,1),(3,2,1,5),(3,2,5,1),(2,2,2,2)$. Send a comment

## Problem