Find all primes $r$ and $s$ such that $ r^7+s=s^3+r $ 
Find all primes $r$ and $s$ such that $\;r^7+s=s^3+r\;.$

I have tried to factorise the expression since we are dealing with primes so:
$$ s(s^2-1)= r(r^6-1)=r(r^2-1)(r^4+r^2+1).$$
Since $ s^3>r^7$, then $\;s^2-1>r^4-1$.
That is all what I have found, thank you in advance for your precious help!

Trying to milk more out of the factorization
$$
s(s-1)(s+1)=s^3-s=r^7-r=r(r-1)(r+1)(r^2-r+1)(r^2+r+1).\qquad(*)
$$
The OP already observed that $s^3>r^7$. Therefore $s>r^2$. Let's remember that $s$ is a prime. Staring at $(*)$ for a few seconds tells us that the only possible prime factor of the RHS larger than $r^2$ is $r^2+r+1$.

Therefore we must have $s=r^2+r+1$.

Plugging this into $(*)$ gives us
$$
\begin{aligned}0&=r^7-r-(r^2+r+1)^3+(r^2+r+1)\\
&=r(r+1)(r-3)(r^2+1)(r^2+r+1).
\end{aligned}
$$
Obviously $r=3$ is the only relevant solution and, luckily, $s=r^2+r+1=13$ turns out to be a prime.
