If $p,q,r$ are all primes,and $p|qr-1$,$q|pr-1$ and $r|pq-1$,find all possible values of $pqr$.

Without loss of generality, assume that $ p \le q \le r.$ Multiplying the three relations gives

$$pqr | p^2q^2r^2 - p^2qr - pq^2r - pqr^2 + pq + pr + qr - 1;$$

therefore $$pqr | pq+pr+qr - 1 < 3qr$$ and thus $p = 2$.

By Hagen's comment, $q$ and $r$ are odd.

Multiplying again gives

$$ qr | (2r - 1)(2q-1) = 4qr - 2q - 2r + 1$$ and therfore $$ qr | 2q + 2r - 1 < 4r,$$ i.e. $ q = 3$.

Finally, the last relation now reads $r | 6 - 1 = 5$ giving $r = 5$.

This means that $pqr = 30$ in all cases.

p, q, r are clearly distinct, suppose that $p>q>r$.

We have $pq|(qr-1)(pr-1)$, which implies $pq|r(p+q)-1$.

Suppose that $2pq\leq r(p+q)-1$, then $2pq\leq r(p+q)-1\leq (q-1)(p+q)-1=pq+q^2-p-q-1$, hence $pq\leq q^2-p-q-1$, contradicts to $p>q$.

Hence, we have $pq=r(p+q)-1$, which implies $r|pq+1$. Together with the assumption $r|pq-1$, we get $r=2$.

Plug into the equation above, we get $(p-2)(q-2)=3$. So $p=5$ and $q=3$.

If $p\mid qr-1$, $q\mid pr-1$ and $r\mid pq-1$, then $$pqr\mid(qr-1)(pr-1)(pq-1).$$ Rewriting the right side, $$pqr\mid (pqr)^2-pqr(p+q+r)+pq+qr+pr-1,$$ so certainly $$pqr\mid pq+qr+pr-1,$$ which implies $pqr\leq pq+qr+pr-1$.

Observe that the RHS will be smaller then the LHS in most cases. This gives many limitations on $p$, $q$ and $r$.