A similar approach as in 2023 RMM/1 should work, I think.

Why one of the primes must be the sum of the other two? What if $q|p+r$ such as $2q=p+r$?

If $p=\max(p,q,r)$ then $p\mid q+r$ implies $p=q+r$ unless $p=q=r$

Why must $\max (p,q,r)|p+q+r$?What if $\min (p,q,r)|p+q+r$?

ok yeah you're right

See that $(p,q,r)=(3,3,3)$ is a solution. Now, we have $p^2qr=p^3+q^3+r^3\ge 3pqr\Rightarrow p\ge 3$. If $p=3$, then we must have equality so $q=r=3$ as well. Assume that $p\ge 5$. One can easily check that if two of the numbers $p,q,r$ are equal, then all of them must be $3$. Thus, $p,q,r$ are pairwise different. We know that $p^2|(q+r)(q^2+r^2-qr)$. Since $p$ doesn't divide $qr$ and $p\neq 3$, we know that $V_p(gcd(q+r,q^2+r^2-rq))=0$. If $p^2|q+r$, then $q+r\ge p^2$. Dividing both sides of the original equation by $p^2$ yields $qr=p+(q^2+r^2-rq)\frac{q+r}{p^2}\ge p+q^2+r^2-qr\ge p+qr$, contradiction. Hence, $p^2|q^2+r^2-rq$. In particular, $p\equiv 1\pmod{3}$. Examine the original equation modulo $3$. As $x^3\equiv_3 x$, we get $1+q+r\equiv_3 qr\Rightarrow (q-1)(r-1)\equiv_3 2$. Hence, either $q$ or $r$ must be $3$. WLOG $r=3$. We get $p^3+q^3+27=3p^2q$. Thus, $q|(p+3)(p^2-3p+9)$. If $q|p+3$, then as $2q=q+q>3+p$, we get $q=p+3$ which means that either $q$ or $p$ is even, which is impossible. Then, $q|p^2-3p+9$. In particular, $q\equiv 1\pmod{3}$. Now we examine the equation $p^3+q^3+27=3p^2q$ modulo $3$ and get a contradiction as $1+1+27\not\equiv 3\pmod 3$.

BarisKoyuncu wrote: In particular, $q\equiv 1\pmod{3}$. Why?

Schur-Schwartz wrote: BarisKoyuncu wrote: In particular, $q\equiv 1\pmod{3}$. Why? This is the only way I know of proving this $$q\mid p^2-3p+9\implies 4p^2-12p+36\equiv (2p-3)^2+27\equiv 0\pmod q$$So $\left(\frac{-27}{q}\right)= \left(\frac{-3}{q}\right) =1$ meaning $q\equiv 1\pmod 3$

It's a special case of the argument that was already used a few lines earlier where $p \mid q^2+r^2-rq$ implies $p \equiv 1 \pmod{3}$. There is no need to invoke quadratic reciprocity here: The element $x=-\frac{q}{r}$ has $p \mid x^2+x+1$ and hence $x^3 \equiv 1 \pmod{p}$ so that the order of $x$ modulo $p$ is $3$. But the order divides $p-1$ and hence $3 \mid p-1$ as desired. This is a special case of the very classical trick that prime divisors of $\frac{x^q-1}{x-1}$ are either $p=q$ or $p \equiv 1 \pmod{q}$ (if $q$ is a prime, in our case $q=3$ so that $\frac{x^3-1}{x-1}=x^2+x+1$).

rightways wrote: Solve the given equation in prime numbers $$p^3+q^3+r^3=p^2qr$$ This problem took me a long time as I didn't think it would be necessary to use the Legendre symbol Solved with complex_afi $\color{blue}\boxed{\textbf{Answer: (3,3,3)}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$p^3+q^3+r^3=p^2qr$$$$\Rightarrow p^2|q^3+r^3$$$$\Rightarrow p^2|(q+r)(q^2-qr+r^2)$$$\color{red}\boxed{\text{If } p^2|q+r :}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow p^3+q^3+r^3=p^2qr\le (q+r)qr$$$$\Rightarrow q^3+r^3<(q+r)qr$$$$\Rightarrow (q+r)(q-r)^2<0 (\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\text{If } p|q+r\text{ and }p|q^2-qr+r^2 :}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow p|q^2-qr+r^2-(q+r)^2$$$$\Rightarrow p|3qr$$$\color{green}\boxed{\text{If } p=3 :}$ $\color{green}\rule{24cm}{0.3pt}$ $$\Rightarrow 27+q^3+r^3=9qr$$By $AM-GM:$ $$\Rightarrow 27+q^3+r^3=9qr\ge 3\sqrt[3]{3^3q^3r^3}=9qr$$$$\Rightarrow q=r=3$$$$\Rightarrow (3,3,3) \text{ is a solution}$$$\color{green}\rule{24cm}{0.3pt}$ $\color{green}\boxed{\text{If } p\neq 3 :}$ $\color{green}\rule{24cm}{0.3pt}$ $$\Rightarrow p|qr$$$\text{WLOG } p=q:$ $$\Rightarrow 2p^3+r^3=p^3r$$$$\Rightarrow p^3|r^3$$$$\Rightarrow p=r$$$$\Rightarrow 3p^3=p^4$$$$\Rightarrow p=3(\Rightarrow \Leftarrow)$$$\color{green}\rule{24cm}{0.3pt}$ $\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\text{If } p^2|q^2-qr+r^2:}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow p^2|4q^2-4qr+4r^2$$$$\Rightarrow p^2|(2q-r)^2+3r^2$$$$\Rightarrow \left( \frac{-3r^2}{p} \right)=1 \text{ or }0\text{(but this gives p=3 or p=r that we already saw in previous cases)}$$$$\Rightarrow \left( \frac{-3}{p} \right)\left( \frac{r^2}{p} \right)=1$$$$\Rightarrow \left( \frac{r^2}{p} \right)=1 \text{ and }\left( \frac{-3}{p} \right)=1$$$$\Rightarrow p\equiv 1\pmod{3}$$$$\Rightarrow 1+q^3+r^3\equiv qr\pmod{3}$$$$\Rightarrow 1+q+r\equiv qr\pmod{3}$$$$\Rightarrow 2\equiv (p-1)(q-1)\pmod{3}$$$$\Rightarrow q\text{ or }r\equiv 0\pmod{3}$$$\text{WLOG } q=3, r\equiv 2\pmod{3}:$ $$\Rightarrow p^3+r^3+27=3p^2r$$$$\Rightarrow r|p^3+27$$$$\Rightarrow r|(p+3)(p^2-3p+9)$$Let's note that $r$ cannot divide $p^3-3p+9$ since analogous to the above we obtain $r\equiv 1\pmod{3}(\Rightarrow \Leftarrow)$ $$\Rightarrow r|p+3$$$$\Rightarrow r\le p+3$$$\text{If }r=p+3 \Rightarrow r\equiv 1\pmod{3}(\Rightarrow \Leftarrow)$ $\text{If }r\le \frac{p+3}{2}:$ $$\Rightarrow p^3<3p^2r$$$$\Rightarrow p<r$$$$\Rightarrow p<\frac{p+3}{2}$$$$\Rightarrow p=2$$$$2^3+q^3+r^3=4qr$$By $AM-GM:$ $$\Rightarrow 4qr\ge 3\sqrt[3]{2^3q^3r^3}=6qr(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow \boxed{(3,3,3)\text{ is the only solution}_\blacksquare}$$$\color{blue}\rule{24cm}{0.3pt}$

Simplest solution- Let p,q,r >5, then p,q,r=6k+-1 Then prove that this isnt possible by just changing the equation then make cases