Find all triples of prime numbers $(p,q,r)$ that satisfy both $p+q+r\mid pq+qr+rp$ and $p+q+r\mid p^3+q^3+r^3-2pqr$

Question

Find all triples of prime numbers $(p,q,r)$ that satisfy: $p+q+r\mid pq+qr+rp$ and $p+q+r\mid p^3+q^3+r^3-2pqr$. I would like you to only check my reasoning. This is an outline. From the problem statement I derived that $p+q+r\mid pqr$. Now we see that $p\mid q+r$, $q\mid r+p$ and $r\mid p+q$. That means $\operatorname{lcm}(p,q,r)\mid p+q+r$. If $p,q,r$ are pairwise different than $\operatorname{lcm}(p,q,r)=pqr$ and hence $\frac{p+q+r}{pqr} = \frac{1}{pq}+\frac{1}{qr}+\frac{1}{rp}$, but its greatest summation element WLOG $\frac{1}{pq}$, $\frac{1}{pq} \le\frac{1}{6}$ hence impossible. Now if $p=r$ we have $pq\mid 2p+q$ but its again impossible. We are left with the case $p=q=r$ and the only pair that satisfies $3p\mid p^3$ is $(p,q,r)=(3,3,3)$.

Answer 1

• How did you derive that $p+q+r$ divides $pqr$?
• How do you see that then $p$ divides $q+r$, and $q$ divides $p+r$, and $r$ divides $p+q$?
• If $p$, $q$ and $r$ are pairwise distinct then $\operatorname{lcm}(p,q,r)=pqr$. Without loss of generality $p<q<r$ and so $$pqr>3r>p+q+r.$$ Your approach is fine too, this just seems more straightforward to me.
• How did you get $3p\mid p^3$ if $p=q=r$?

P.S. I can answer each of the questions above myself, but your reasoning is missing.