Find all prime numbers $p$, $q$ and $r$ such that the fourth power of any of them minus one is divisible by the product of the other two. (Author: V. Senderov)
$ p^{4}-1=qra_{1} $
$ q^{4}-1=pra_{2} $
$ r^{4}-1=qpa_{3} $.
let take $ p $ the minimum.
$ (p-1)(p+1)(p^{2}+1) $ is divisible by $ qr $.
so $ (p+1)(p^{2}+1) $ is divisible by $ qr $ so $ p+1 $ is divisible by $ q $ because $ p^{2}+1<qr $.
so $ p+1<q+1 $ and that means $ p+1=q $ so $ p=2,q=3,r=5 $ and the cyclic permutations.