Case $ p, q$ and $ r$ are odd, we have $ p^2p \equiv p^2 + q^2 + r^2 mod 4 \Rightarrow 1p \equiv 1 + 1 + 1 \equiv 3 mod 4$. But $ p|q^2 + r^2$, which implies $ p|q, r$ if $ p = 4k + 3$. Then $ p = q = r$. Now we just need parity arguments. Answer: $ p = q = r = 2$.

feliz wrote: Answer: $ p = q = r = 2$. Are you sure?

The first if $ p=3$ we have $ (3,3,3)$ is a solution Now if $ p>3$ If $ q,r\#3$ so $ p^2+q^2+r^2$ divides to $ 3$ thus $ p$ divides to $ 3$ it is contraction Hence W.L.G we can suppose that $ q=3$ If $ r=3$ so $ p=3$ If $ r>3$ Our equation becomes $ p^2(p-1)=r^2+9$ Because $ p$ is a prime so we must have $ p-1$ divides to $ r^2+9$ or $ p=r^2+9$ And they follow us that $ p^2(p-1)>r^2+9$ We can finish our solution and the answer is $ (3,3,3)$ Am I wrong ?????

feliz wrote: Case $ p, q$ and $ r$ are odd, we have $ p^2p \equiv p^2 + q^2 + r^2 mod 4 \Rightarrow 1p \equiv 1 + 1 + 1 \equiv 3 mod 4$. But $ p|q^2 + r^2$, which implies $ p|q, r$ if $ p = 4k + 3$. Then $ p = q = r$. Now we just need parity arguments. Answer: $ p = q = r = 2$. a mistake.if p=q=r we must i.e p^3=3p^2 so p=q=r=3. a application of Fermat-euler theorem

$\pmod 3$ implies that one of $p,q,r=3$ Suppose that $p=3$ then $18=q^2+r^2$ with $(q,r)=(3,3)$ Now if either $q$ or $r=3$ then $p^3-p^2=q^2+9$ By $\pmod 4$ ,$p\equiv 3 \pmod 4$ Choose a prime divisor $p_i|p^3-p^2$ which is $\equiv 3 \pmod 4$ Since if $p|a^2+b^2$ with $p \equiv 3 \pmod 4$ it implies $p|a$ and $p|b$,and we are done since this implies $p^3-p^2=18$ or $p=3$ which is same as before so the only solution is $(p,q,r)=(3,3,3)$

The answer is $(3, 3, 3)$ only. Notice that if $3 \not \in \{p, q, r\}$, then we have mod $3$ issues. Now if $p=3$, then $(q, r) = (3, 3)$ obviously. Thus let $ p > 3$ and $r = 3$, or $$p^2(p-1) = q^2 + 9.$$For mod $4$ issues $p \equiv 3 \pmod 4$, then $p \mid q^2+9$ and by Fermat Christmas $p \mid q$, so we get $(3, 3, 3)$ again.