# Can the sum of powers of the first primes be a square?

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.

• Did you find more solutions when $u=1$? – vrugtehagel Apr 3 '17 at 12:12
• 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 3 '17 at 13:29
• OEIS A061888 gives 10, 5063, 14573, 17098, 1916357, 468726713734 – Ross Millikan Apr 8 '17 at 18:34
• 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 9 '17 at 3:22
• 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 9 '17 at 3:32