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Let $q$ be an odd prime and $s\ge 2$ be an integer. Define the sum

$$S(q,s):=\sum_{p\ prime,p\le q} p^s=2^s+3^s+\cdots +q^s$$

Can $S(q,s)$ be a perfect power ? Among other searches, I searched for $s\le 1000$ and $q\le1000$ and did not find an example.

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  • $\begingroup$ @Ethan I did various searches. For the exponents upto $30$, I checked upto $q\le 10^7$ $\endgroup$ – Peter Apr 28 '18 at 21:26
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    $\begingroup$ For exponents $2\le s\le 120$ , $q$ must be greater than $10^6$ , for $2\le s\le 10$, $q$ must be greater than $10^8$ $\endgroup$ – Peter Apr 28 '18 at 21:44
  • $\begingroup$ Just to do feedback about it, but I think that my comment isn't useful and has no a good mathematical content (I'm agree that my comment is fantasy): if we assume that we can split (case of an even number $2L$ of summands in your finite series) your expression as $L$ pairs of expressions $p^s+(\hat{p})^s=\frac{1}{L}x^y$, where also we assume that $L\mid x$, the abc conjecture implies that $\forall\epsilon>0$ does exist a constant $C(\epsilon)$ such that $\frac{1}{L}x^y\leq C(\epsilon)(p\cdot \hat{p}\cdot\frac{x}{L})^{1+\epsilon}$. $\endgroup$ – user243301 Apr 28 '18 at 21:45
  • $\begingroup$ For $s=2$ and $s=3$, $q$ must be greater than $3\cdot 10^9$ $\endgroup$ – Peter Apr 28 '18 at 21:53
  • $\begingroup$ In the case $s=2$ , $q$ must be greater than $11\cdot 10^9$ $\endgroup$ – Peter Apr 29 '18 at 8:33

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