Prime Power

$r \geqslant 2$ $\Rightarrow$ $p>q$

First : If $q>2$

$q^{p}$ is odd & $p^{q}$ is odd too $\rightarrow$ $\frac{q^{p}+p^{q}}{2}\in N$ $\rightarrow$ r=2 $\rightarrow$ $p^{q}+q^{p}=2^{p-q}$ but $q^{p} >2^{p-q}$ $\Rightarrow$ $p^{q}+q^{p} > 2^{p-q}$

second : If $q=2$

• $p=3$ $\Rightarrow$ $\boxed{r=17}$

• $p>3$ then p= $n \pm 1$ (n: $\frac {n}{3}\in N$ ) , $p^{2}+2^{p}$ = $(n\pm1)^{2}+2^{p}$ = $n^{2}\pm2n+1+2^{p}$ , $\frac {2^{p}+1}{3} \in N$ ( $p$ is prime greater than 2 ) $\rightarrow$ $\frac{n^{2}\pm2n+1+2^{p}}{3}\in N$ $\rightarrow$ $r=3$ this gives two statements:

  1 . $p=5$ $\Rightarrow$ $5^{2}+2^{5}\neq3^{5-2}$

  2 . $p>5$ $\Rightarrow$ $p^{2}+2^{p}<3^{p-2}$

$\therefore$ Hence: $p=3,q=2$ and $r=17$ is the only solution

The left side is powered to two numbers while the right is powered to their difference; q and p ramp up much faster than p-q, so r gets bigger the bigger numbers you use

You're welcome to check, but I just didn't include that in my short solution. :D

Yeah. :D