Playing with the primes

Write $za = xy-1$ ; then $a$ is positive. Note that $a \equiv -x$ mod $y$ and $a \equiv -y$ mod $x$ (by the other two divisibility conditions). For now let's suppose $x < y < z$ . Then $0 < a < \frac{xy}{z} < y$ , and we know what $a$ is mod $y$ , so there is only one possibility, namely $a = y-x$ . But then we get $y-x \equiv -y$ mod $x$ so $2y \equiv 0$ mod $x$ . Since $x$ and $y$ are distinct primes, this implies $x = 2$ .

So now we have $z(y-2) = 2y-1$ ; rearrange to get $(y-2)(z-2) = 3$ , whose only solution with $y < z$ is $(3,5)$ . So the unique solution with $x < y < z$ is $(2,3,5)$ . Permuting this solution to get all possibilities yields $\fbox{6}$ total solutions.