find primes p,q,and r 
Find all primes $p, q, r$ such that $pq+r$, $pq+r^2$, $qr+p$, $qr+p^2$, $rp+q$,
$rp+q^2$ are all primes.


We know that $pq+r>2$. If all of $p,q,r$ are odd, then $pq+r$ would be even, which would make is impossible for the number to be prime. Now, without loss of generality, let $p=2$. We then have $qr+2$ and $qr+4$ to be prime. We know that both these numbers are greater than $3$ and prime, thus cannot be divisible by $3$.
If $qr$ leaves remainder $1$ modulo $3$, then $qr+2$ would be divisible by $3$. If $qr$ leaves remainder $2$ modulo $3$, then $qr+4$ would be divisible by $3$. Thus, we must have $qr$ divisible by $3$. Without loss of generality, let $q=3$. Now, we need the following numbers to be prime:
$$r+6,r^2+6,2r+3,2r+9,3r+2,3r+4$$
All of the above primes must be greater than $5$, and thus, none of them are divisible by $5$. Consider the possible remainders when $r$ is divided by $5$. If $r \equiv 1 \pmod{5}$, then $2r+3$ would be divisible by $5$. If $r \equiv 3 \pmod{5}$, then $2r+9$ would be divisible by $5$. If $r \equiv 2 \pmod{5}$, then $3r+4$ would be divisible by $5$. Finally, if $r \equiv 4 \pmod{5}$, then $r+6$ would be divisible by $5$. As none of these are the case, $r$ is divisible by $5$, and thus, $r=5$. We can verify that this solution works.
Thus, the possible triples are:
$$(p,q,r)=(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2)$$
