I want to know properties of the following sum: $$\sum_{j=0}^{p-1} \omega^{\beta j^2}= ~? $$ where $p$ is a prime, and $\omega^p=1$, is a $p$th root of unity (and $\beta$ is an integer between $0$ and $p-1$). If anyone has any references related to this, let me know.

Also, if someone can tell me about the sum: $$\sum_{j=0}^{p-1} \omega^{\beta j^n}= ~? $$ for higher values of $n$ such as 3, that would also be nice.

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    $\begingroup$ What are the $\beta_j$'s? $\endgroup$ Jan 16, 2018 at 8:59
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    $\begingroup$ @DesideriusSeverus I believe it is $\beta j$, not $\beta_j$, so $\beta$ is just some parameter. $\endgroup$ Jan 16, 2018 at 9:08
  • $\begingroup$ @BrendanMcKay Indeed, I should wear glasses! My question could be interpreted then as "is there any information on $\beta$? $\endgroup$ Jan 16, 2018 at 9:12
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    $\begingroup$ Thanks @Nemo for the links. I found the wikipedia article a bit later myself after some searching. $\endgroup$
    – guest17
    Jan 16, 2018 at 9:56

2 Answers 2


Your first sum is a special Gauss sum. For its value in general, see Corollary 9.16 in Montgomery-Vaughan: Multiplicative number theory I. Your second sum can also be expressed in terms of Gauss sums (associated with primitive characters modulo $p$), and in particular its absolute value is at most $(n-1)\sqrt{p}$, assuming $p\nmid\beta$. See Exercise 4c in Section 9.2.1 of the same book.

  • $\begingroup$ Hi,Thanks for the help; for $n=2$ the answer is now very explicit. But I was wondering if there is any simple result for the case when $n=3$. For example, if $p=3,~5,~11,~17$ the sum is equal to zero. So can anyone tell me the values of $p$ for which the sum is zero? I am sure there is a reasonably explicit closed form for the sum when $n=3$ at least. $\endgroup$
    – guest17
    Mar 7, 2019 at 10:26
  • $\begingroup$ @guest18: I don't think there is a simple closed form (even for $n=3$). On the other hand, it is easy to show (for general $n$ and $p\nmid\beta$) that the sum is zero if and only if $n$ is coprime to $p-1$. $\endgroup$
    – GH from MO
    Mar 7, 2019 at 11:05

Experiment makes it clear that this sum is $\epsilon\delta\sqrt{p}$ where

  • $\epsilon=1$ if $p=1\pmod{4}$, and $\epsilon=i$ if $p=-1\pmod{4}$
  • $\delta=1$ if $\beta$ is a quadratic residue mod $p$, and $\delta=-1$ otherwise.

I have not tried to prove this, but I expect that number theorists will not find it hard.

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    $\begingroup$ We will not find it hard indeed because it is well known, but it took Gauss four years... $\endgroup$
    – Olivier
    Jan 16, 2018 at 12:07

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