20
$\begingroup$

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.

$\endgroup$
8
  • 12
    $\begingroup$ 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? $\endgroup$ Mar 5 '18 at 13:42
  • 3
    $\begingroup$ 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. $\endgroup$ Mar 5 '18 at 22:00
  • 3
    $\begingroup$ oeis.org/A031971 (obtained plugging 4 terms 1, 5, 36, 354) $\endgroup$
    – YCor
    Mar 5 '18 at 22:53
  • 2
    $\begingroup$ I checked all $n \leq 4000$, and there were no new examples. $\endgroup$
    – S. Carnahan
    Mar 7 '18 at 14:52
  • 3
    $\begingroup$ 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 $3 \leq p \leq 149$. $\endgroup$ Mar 8 '18 at 18:16
20
$\begingroup$

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 $\sum_{\mathbb N} n^{-n/2} \approx 1.7788$.

$\endgroup$
6
  • 3
    $\begingroup$ 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. $\endgroup$
    – Wojowu
    Mar 5 '18 at 17:41
  • 15
    $\begingroup$ 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. $\endgroup$ Mar 5 '18 at 18:41
  • 3
    $\begingroup$ 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 ? $\endgroup$ Mar 6 '18 at 0:44
  • $\begingroup$ (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.) $\endgroup$ Mar 7 '18 at 18:08
  • 1
    $\begingroup$ (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). $\endgroup$ Mar 7 '18 at 18:24
9
$\begingroup$

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 $8k+4$).

$\endgroup$
7
  • 2
    $\begingroup$ 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. $\endgroup$ Mar 5 '18 at 19:16
  • 1
    $\begingroup$ Exact, indeed $S(8k+6)=3\mod 4$ for all $k$, by a similar computation. $\endgroup$ Mar 5 '18 at 19:18
  • 1
    $\begingroup$ "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) $\endgroup$ Mar 6 '18 at 0:23
  • 1
    $\begingroup$ @PietroMajer no, only eventually periodic. For instance $\varphi(8)8^2=256$, but modulo 8 we have $S(2)\equiv 5$ while $S(258)\equiv 1$. $\endgroup$
    – YCor
    Mar 7 '18 at 22:58
  • 1
    $\begingroup$ We may exclude also odd multiples of $4^m$, see my comment to the question $\endgroup$ Mar 10 '18 at 8:02
1
$\begingroup$

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.


$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.