Let $(a,b,c,d)$ be a quadruple of natural numbers, all greater than $1$, such that $d\mid abc+1$, $c\mid abd+1$, $b\mid acd + 1$ and $a\mid bcd+1$. Let's suppose that $a\leq b\leq c\leq d$. Find all such quadruples.
It is clear that these numbers should be coprime (pairwise), since if $k>1$ is such that, say, $k\mid a$ and $k\mid b$, then $a\mid bcd+1$ implies $k\mid bcd+1$, but $bcd+1$ and $k$ are coprime. In particular numbers are different. It seems that there are no such quadruples other than $(2,3,7,43)$, but how does one show that? I don't see any argument, that would somehow use all this $t\mid xyz+1$ to get the unique solution... It is interesting, that this solution involves only primes and that $d$ is exactly $abc+1$. I can see a solution if one proves that $d$ must be equal to $abc+1$, but still, I have no idea how to do that. Can you give any hint (please, not a full solution)?