Fix a prime $p >2$ and $q_1$, $q_2$ such that $q_i - 1$ is exactly divisible by $p$. For any $n$, $a$, $b $, consider the sum


Is this always divisible by $p^{n-1}$? In fact, perhaps it is always $0$ or all the summands are equal? I believe the following question is also relevant. For any j, is

$$\sum_{i=0}^{p^{n-1}-1}\zeta_{p^n}^{pij+bq_2^i} \equiv 0 \pmod{p^{n-1}}? $$

(If $q_1 =q_2$, I think this is true and not hard to see. I am really interested in a more general version but this is the easiest case I don't know how to do.)

  • 2
    $\begingroup$ Does "exactly divisible by $p$" mean "divisible by $p$ but not by $p^2$" here? $\endgroup$
    – LSpice
    May 18 at 23:17
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    $\begingroup$ Yes. This is not an important constraint, I only chose it to make thing explicit. In general the sum should be large enough so that you cover an entire period for $(q_1^i,q_2^i)$. $\endgroup$
    – Asvin
    May 18 at 23:26
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    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    May 19 at 0:35

1 Answer 1


Using the sagemath code,

p = 5
q1 = p+1
q2 = 2*p+1
n = 3
a = 3
b = 1
k = CyclotomicField(p^n)
s = sum([k.gen()^(a*q1^i+b*q2^i) for i in range(0,p^(n-1))])

it output 1/88817841970012523233890533447265625. Hence, the answer is no.

  • 1
    $\begingroup$ For this example Mathematica gives: $zeta = Exp[2 Pi I/125]; \sum _{i=0}^{24} \text{zeta}^{3\ 6^i+11^i}$ gives $$5 e^{\frac{8 i \pi }{125}}+ 10 e^{-\frac{1}{125} (42 i \pi )}+10 e^{\frac{58 i \pi }{125}}$$, which gives N[\%] 10.9551 +2.23283 i. With $a=1$ it is numerically close to $0$. Can anybody else check this independently? $\endgroup$ May 21 at 6:00

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