<s>[INST] 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!
 [/INST] This isn't a full-fledged answer yet but it is too long for a comment: On the one hand,
$$r|(s^2-1) \Rightarrow s=kr \pm 1$$ $$\Rightarrow s \equiv_r \pm 1.$$
On the other hand, $s \in \theta(r^{\frac{7}{3}})$ and $s$ divides one of $r^3-1$, $r^3+1$ gives $as = r^3\pm 1$ for some $a \in O(r^{3-\frac{7}{3}}) =O(r^{\frac{2}{3}})$.
But $as \equiv_r \pm a$ because $s \equiv_r \pm 1$, so either $a=1$ [which is impossible] or $a \ge r-1$. But both $a \in O(r^{\frac{2}{3}})$ and $a\ge r-1$ is impossible for $r$ sufficiently large. So the solution set is bounded.
ETA: You could also use the fact that both $r^3-1$ and $r^3+1$ factor further to polynomials no larger than $r^2+r+1$ [as noted in @John Omielan 's answer below, I had forgotten this], to conclude that $s \le r^2+r+1$, because $r,s$ prime $\Rightarrow$ $s$ must divide one of $r^2+r+1$, $r^2-r+1$, $r-1$, $r+1$. Then this gives $r^7-r \le (r^2+r+1)^3-(r^2+r+1)$ [because $y^3-y$ is strictly increasing in $y$ for $y \ge 3$]. All solutions for $r^7-r \le (r^2+r+1)^3-(r^2+r+1)$ require $r \le 13$. This leaves very few possibilities to check, check the value of $s$ that satisfies $s^3-s=r^7-r$ for each $r \in \{3,5,7,11,13\}$.
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