# Has any one seen this sum of roots of unity before?

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

$$\sum_{i=0}^{p^{n-1}-1}\zeta_{p^n}^{aq_1^i+bq_2^i}.$$

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.)

• Does "exactly divisible by $p$" mean "divisible by $p$ but not by $p^2$" here? May 18 at 23:17
• 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 $(q_1^i,q_2^i)$. May 18 at 23:26
• Please use a high-level tag like "nt.number-theory". I added this tag now. May 19 at 0:35

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))])
(s/p^(n-1)).norm()


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

• 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? May 21 at 6:00