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Let $p$ be a prime and $u\ge 1$ be a positive integer.

Define $$\begin{align} S(p,u) &:= \sum_{q\text{ prime, }q \le p} q^u \\ &= 2^u+3^u+\cdots +p^u\end{align}$$

I wonder whether $S(p,u)$ can be a square for $p>2$ and $u>1$. For $p=2$, we simply have a power of $2$, which is a square , if $u$ is even. For $u=1$, we have $$2+3+\cdots 23=100=10^2.$$

But are there any squares for $p>2$ and $u>1$ ?

I checked the range $2\le u\le 100$ and $3\le p\le 10^8$ without finding a square.

ShreevatsaR
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Peter
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  • Did you find more solutions when $u=1$? –  Apr 03 '17 at 12:12
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    Yes, If we denote $$S_p:=2+3+5+\cdots p$$ we have $S_{23}=10^2$ , $S_{22073}=5063^2$ , $S_{67187}=14573^2$ , $S_{79427}=17098^2$ , $S_{10729219}=1916357^2$ and no other perfect powers (in particular, no further square) for $p\le 10^9$ – Peter Apr 03 '17 at 13:29
  • No, I meant numerical values. To see if the one you gave could be the only one –  Apr 08 '17 at 18:29
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    OEIS A061888 gives 10, 5063, 14573, 17098, 1916357, 468726713734 – Ross Millikan Apr 08 '17 at 18:34
  • @RossMillikan Thanks, so I need not continue my search :) – Peter Apr 08 '17 at 18:35
  • You might be interested in this: http://math.stackexchange.com/questions/908618/is-sum-of-square-of-primes-a-square-of-prime – Fabian Schn. Apr 08 '17 at 18:37
  • $468726713734^{\ 2}$ has $24$ digits. How on earth could the first $m$ primes giving this sum be calculated ? – Peter Apr 08 '17 at 18:38
  • Using a computer with a buttload of memory? Also, you already answered my question before you asked for clarification –  Apr 08 '17 at 18:42
  • @vrugtehagel Or some trick like as in the calculation of the number of primes below $10^{24}$ ? – Peter Apr 08 '17 at 18:43
  • This would make sense had you not deleted your comment. Never do that, especially not when people responded to it –  Apr 08 '17 at 20:03
  • I don't know why you're trying to hide it. You deleted the comment between the second and the now third comment. You asked for clarification on what I meant. I find it disrespectful that you handle the situation this way. –  Apr 08 '17 at 20:49
  • Now I think I know what you mean. I added the information to my question, and then the comment became obsolete. To avoid unnecessary duplication, I deleted it. So, I do not hide anything. And deleting obsolete comments is not disrespectful to my opinion. What you think is missing is the range I checked, and this range is given in my question. – Peter Apr 08 '17 at 20:54
  • Suppose $S(p,u)$ are "random" number and the probability of them being a square is $O(1/\sqrt{S(p,u)})$. It is known that $S(p,u) = O(p^{u+1}/\ln p)$. Therefore, $O(1/\sqrt{S(p,u)})=O(\sqrt{\ln p}/p^{(u+1)/2})$. Since $\sqrt{\ln p}/p^{(u+1)/2}>1/p$ for $u=2$ and $1/p$ diverges, heuristically there are likely to be infinitely many squares among $S(p,2)$ and finitely many squares if $u>2$. – didgogns Apr 09 '17 at 03:15
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    However, one should not take heuristic argument seriously, for example, there are only few integer solution to $\sum_{i=1}^n i^2=m^2$ ($n=24$ is largest solution and it was proved) and infinitely many solution to $\sum_{i=1}^n i^3=m^2$, which is opposite result from what heuristic argument gives. – didgogns Apr 09 '17 at 03:22
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    Oh, bad me, $\sqrt{\ln p}/p^{(u+1)/2}$ converges for $u \ge 2$ because it is smaller than p-series of $p=1.5-\epsilon$. Heuristically there are likely to be finitely many solutions for $u>1$. – didgogns Apr 09 '17 at 03:32

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