If $p,q,r$ are all primes,and $p|qr-1$,$q|pr-1$ and $r|pq-1$,find all possible values of $pqr$. If $p,q,r$ are all primes,and
$p|qr-1$$~~~~~~~~~~~~~$$q|pr-1$$~~~~~~~~~~$ and $~r|pq-1$.
Find all possible values of $pqr$.  
My work:
$qr-1=pk_1$
$pr-1=qk_2$
$pq-1=rk_3$
From the above equations, we can conclude that either $k_1,k_2,k_3$ are all even or one of the primes is $2$.
I can also get that,
$p^2q^2r^2=(pk_1+1)(qk_2+1)(rk_3+1)$
Now, I cannot proceed. Please help!
 A: Without loss of generality, assume that $ p \le q \le r.$
Multiplying the three relations gives
$$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,$$ i.e. $ q = 3$.
Finally, the last relation now reads $r | 6 - 1 = 5$ giving $r = 5$.
This means that $pqr = 30$ in all cases.
A: If $p\mid qr-1$, $q\mid pr-1$ and $r\mid pq-1$, then $$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 $r$.
A: 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 $q=3$.
A: Wlog $\, p<q<r,\ $ else, $ $ e.g. $\ q=p\mid qr\!-\!1\, \Rightarrow\, q\mid 1\Rightarrow\Leftarrow$   
$p,q,r\mid n=qr\!+\!pr\!+\!pq\!-\!1 \,\Rightarrow\, \color{#c00}pqr\mid n<\color{#c00}3qr \,\Rightarrow\, \color{#c00}{p=2}$   
$\ \ \ q,r\mid n\ \ \ \,=\,\ \ \ \color{#c00}2r\!+\!\color{#c00}2q\!-\!1\ \,\Rightarrow\,\  \color{#0a0}qr\mid n\, <\, \color{#0a0}4r\ \Rightarrow\  \color{#0a0}{q=3}$    
$\quad\ \ r\mid pq\!-\!1=5\,\Rightarrow\, r=5\ \ \ $  QED
