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

Solve the system in natural numbers

\[
\begin{aligned}
& x+y=z t \\
& z+t=x y .
\end{aligned}
\]
Consider two cases.

1) $x=1$. Then $z+t=y=z t-1$, hence $(z-1)(t-1)=z t-z-t+1=2$. 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, 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

## [ Numerical tables and their properties ]  Problem 78651 Topics: [ Examples and counterexamples. Constructions] [ Pairings and groupings; bijections ]

Arrange numbers in a $4 \times 416$ table so that the sum of the numbers in any vertical, horizontal, and diagonal line equals zero. (The table has 14 diagonals, including all small ones consisting of three, two, and one cells. At least one of the numbers must be different from zero.)

## Answer

| 0 | $a$ | $-a$ | 0 |
| :---: | :---: | :---: | :---: |
| $-a$ | 0 | 0 | $a$ |
| $\mathrm{a}$ | 0 | 0 | $-a$ |
| 0 | $-a$ | $a$ | 0 |

Given two natural numbers $m$ and $n$. All different divisors of the number $m-$ numbers $a, b, \ldots, k-$ and all different divisors of the number $n$ - numbers $s, t, \ldots, z$. (The number itself and 1 are also included in the divisors.) It turned out that $a+b+\ldots+k=s+t+\ldots+z$ and $1 / a+1 / b+\ldots+1 / k=1 / s+1 / t+\ldots+1 / z$.

Prove that $m=n$.

## Solution

If $d$ is a divisor of the number $n$, then ${ }^{n / d} d$ is also a divisor of the number $n$. Therefore, the sets of numbers $(a, b, \ldots, k)$ and ( ${ }^{n} / a,{ }^{n} / b$, $\ldots, n / k$ ) coincide, from which

$n / a+n / b+\ldots+n / k=a+b+\ldots+k$

Similarly, $m / s+m / t+\ldots+m / z=s+t+\ldots+z$.

Dividing one of the obtained equalities by the other, we get $n=m$, which was to be proved.