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.)



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


