# When is $1^n+2^n+3^n+4^n+\cdots+n^n$ a square number?

I've been thinking about such an interesting question these last few days. I wonder if anyone has studied it.

Question: Find all postive integers $n$ such that $$S(n)=1^n+2^n+3^n+4^n+\cdots+n^n$$ is a square number?

Here $n=1$ is clear and if $n=3$ then $S(3)=36=6^2$. Now I can't find any other $n\le 20$, so I conjecture these $n$ are the only two values.

• What have you done so far? Did you predict how often this should happen based on asymptotics? Check for obstructions (mod n)? Plug the sequence into OEIS? Mar 5, 2018 at 13:42
• If $n=2^k s$ and $k>0$ is even, then $S(n)$ is divisible by $2^{k-1}$, but not by $2^k$, thus can not be a perfect square. Analogously, if $n=4k+2$, then $S(n)$ is congruent to $n/2=2k+1$ modulo 8, thus $k$ must be divisible by 4. Mar 5, 2018 at 22:00
• oeis.org/A031971 (obtained plugging 4 terms 1, 5, 36, 354)
– YCor
Mar 5, 2018 at 22:53
• I checked all $n \leq 4000$, and there were no new examples. Mar 7, 2018 at 14:52
• By computing $S(n)$ efficiently mod $p$ for small $p$, I have verified that $S(n)$ is not a square for $4 \leq n \leq 10^{7}$. The most difficult $n$ was $n = 9676659$ for which $S(n)$ is a square mod $p$ for $3 \leq p \leq 149$. Mar 8, 2018 at 18:16

This is not an answer, but I don't have enough reputation to comment. For large $n$, $S(n)\approx n^n.$ The density of square integers at $n^n$ is approximately $n^{-n/2}$. So one might expect that the number of $S(n)$ which are square is about $\sum_{\mathbb N} n^{-n/2} \approx 1.7788$.

• I believe $S(n)$ is better approximated by $\frac{e}{e-1}n^n$, but up to a constant the order of magnitude is right. Mar 5, 2018 at 17:41
• And given that the OP checked up to $n=20$ and the fact that $\sum_{n\ge21} n^{-n/2}\approx 1.5\cdot 10^{-14}$, the chances of there being any other examples is vanishingly small. Mar 5, 2018 at 18:41
• Denoting by $C=\dfrac{e}{e-1}$, one has $C^{-1}(\sum_{n>0}n^{-n/2})^{2}\approx 2$ . Is there a reason for this ? Mar 6, 2018 at 0:44
• (Technical comment to take new developments into account: two commenters at 2018-03-07 14:52:58Z and 2018-03-07 16:30:52Z not only "checked up to $n=20$ but up to $n=5500$, and still did not find another example.) Mar 7, 2018 at 18:08
• (but the argument also applies to the sequence $S_0(n){\bf =}n^n$, which is nevertheless a square more than half of the time). Mar 7, 2018 at 18:24

An easy obstruction (mod 8). We can exclude all odd multiples of $4$, because for any $k\in\mathbb{N}$, $S(8k+4)=2\mod 4.$ (Reason: if $j$ is even, $j^{8k+4}=0\mod 4$, and if $j$ is odd, $j^{8k+4}=1\mod 4$, and there are $4k+2$ odd numbers from $1$ to $8k+4$).

• You can also exclude those values which lead mod 8 to anything other than 0,1, or 4. This takes out a few more numbers mod 16. Gerhard "Then Try Mod Three Next" Paseman, 2018.03.05. Mar 5, 2018 at 19:16
• Exact, indeed $S(8k+6)=3\mod 4$ for all $k$, by a similar computation. Mar 5, 2018 at 19:18
• "Then Try Mod Three Next" : this takes out 3 numbers mod 18. In fact, for any prime $p$, $S(n)\mod p$ is periodic with period $(p-1)p^2$. For p=5, this takes out 15 numbers mod 100. (Not sure if this way one can exhaust all n though) Mar 6, 2018 at 0:23
• @PietroMajer no, only eventually periodic. For instance $\varphi(8)8^2=256$, but modulo 8 we have $S(2)\equiv 5$ while $S(258)\equiv 1$.
– YCor
Mar 7, 2018 at 22:58
• We may exclude also odd multiples of $4^m$, see my comment to the question Mar 10, 2018 at 8:02

A special case,

Thm 1:   if prime $\ p\equiv 3\$ mod $4,\$ then $\ 1^{p-1}+\ldots (p-1)^{p-1}\$ is not a square.

Not a real answer either, but this link may be relevant : Faulhaber's formula

In particular one could try to establish, following Pietro Majer and Gerhard Paseman, that a solution of the diophantine equation $S(n)=m^{2} =P_{n}(a)$ only occurs for odd values of $n$, where $a=n(n+1)/2$ and $P_{n}$ is the $n$-th Faulhaber polynomial.