Mitkin D:

Find all quadruples of real numbers, in each of which any number is equal to the product of any two other numbers.
$0,0,0,0,1,1,1,1,-1,-1,1,1,-1,-1,-1,1$ (up to permutation of the numbers in the quartet). Consider the absolute values of the desired numbers and order them in non-decreasing order: $a \leq b \leq c \leq d$. Note that $a \geq b c$, since either $a=b c$, or $a=b d \geq b c$, or $a=c d \geq b c$. Similarly, $d \leq b c$. Therefore, $b c \leq a \leq b \leq c \leq d \leq b c$, i.e., $a=b=c=d=b c=x$. Since $x=x^{2}$, then $x=0$ or $x=1$. It remains to check that if there are negative numbers in the quartet, their number is two or three.

## Answer

$0,0,0,0,1,1,1,1,-1,-1,1,1,-1,-1,-1,1$ (up to permutation of the numbers in the quartet).