Tintarn wrote: Determine all real numbers $a,b,c$ and $d$ with the following property: The numbers $a$ and $b$ are distinct roots of $2x^2-3cx+8d$ and the numbers $c$ and $d$ are distinct roots of $2x^2-3ax+8b$. We have system : $2a+2b=3c$, $ab=4d$, $2c+2d=3a$, $cd=4b$ $\iff$ $d=\frac{ab}4$ and $2a+2b=3c$, $4c+ab=6a$, $abc=16b$ $a=0$ or $b=0$ would imply $a=b=c=d=0$, discarded. So $ab\ne 0$ and problem is : $d=\frac{ab}4$, $c=\frac{16}a$ and $a^2+ab=24$, $64+a^2b=6a^2$ Cancelling $b$ from two last equalities, we get $a^3+6a^2-24a-64=0$ Which is $(a-4)(a+2)(a+8)=0$ And solutions $\boxed{(a,b,c,d)\in\{(4,2,4,2),(-2,-10,-8,5),(-8,5,-2,-10)\}}$

This is exactly the same as problem 8.4.4 from AoPS's Intermediate Algebra, which cites the problem as previously being from Costa Rica (though I cannot find the specific source on AoPS).