How do I prove that $x^s=(-1)^k \sum_{k=0}^{(r-1)/2}\binom{r}{2k}p^{s(r-2k)}$ has no solutions?

Question

I have been struggling to prove that the following diophantine equation has no integral solutions if $r$ is odd, $s,p>1$ $$x^s=(-1)^k \sum_{k=0}^{(r-1)/2}\binom{r}{2k}p^{s(r-2k)}$$ Any hint on how to approach this beast

Answer 1

If $r=1$, this becomes $x^s = p^s$ which has the solution $x = p$.

Answer 2

If $r=3$, this is $x^s =\binom{r}{0}p^{3s}-\binom{r}{2}p^{s} = p^{3s}-\frac{r(r-1)}{2}p^s = p^s(p^{2s}-3) $.

By considering that all the prime factors of $x$ and $p$ are divisible by $s$, this must also hold for $p^{2s}-3$, so $p^{2s}-3 = d^s$ for some $d$.

Therefore $d^s = p^{2s}-3 $ or $3 =p^{2s}-d^s $.

$3 =p^{2s}-d^s \ge (d+1)^s-d^s \ge sd^{s-1} $, so $s \le 3$. If $s = 3$, $(d+1)^3-d^3= 3$ which has no solutions. If $s=2$, $3 \ge (d+1)^2-d^2 =2d+1 $ so $d=1$ and $3=p^4-1$ which again has no solutions.

Therfore this has no solution for $r = 3$.