Let $p$ be an integer polynomial and $k$ be a natural number. Is the exponential sum

$$C(\alpha) = \sum_{i=1}^k e(\alpha \{p(i)/k\})$$

identically 0 only if $\alpha$ is an integer? Here, $e(x) = e^{2 \pi i x}$ is the usual complex exponential and $\{x\}\in[0,1)$ is the fractional part of a real number $x$.

As strong as this conjecture may sound, it has been confirmed by all the examples that I have checked. For instance, if $p(i) = i$ and $k=4$, then

$$C(\alpha) = 1 + e(\alpha/4) + e(2\alpha/4) + e(3\alpha/4) = (1+e(\alpha/4))(1+e(\alpha/2))$$

is 0 if and only if $\alpha = 1,2,3 + 4\mathbb{Z}$.

Here is a potentially useful reformulation. Let $q$ be an integer polynomial for which $C(\alpha) = q(e(\alpha/k))$: in the example above, this is $$q(z) = 1 + z + z^2 + z^3 = (1+z)(1+z^2).$$ Then the conjecture is tantamount to showing that any root of $q$ on the unit circle is necessarily a $k$-th root of unity.

Note that the conjecture fails if we replace the set $\{p(1), \ldots, p(k)\}$ by arbitrary $b_1, \ldots, b_k\in\{0, \ldots, k-1\}$. For instance, if $(a,b,c)$ is a coprime Pythagorean triple and $$ q(z) = c \left(z + \frac{a}{c} + \frac{b}{c}i\right)\left(z + \frac{a}{c} - \frac{b}{c}i\right) = cz^2 + 2a z + c,$$ then this polynomial has roots on the unit circle that are not roots of unity, yet it corresponds to the sum $$ C(\alpha) = \sum_{i=1}^k e(\alpha b_i/k)$$ where $k=2(a+c)$ and $b_i$'s take values 0, 1 and 2 for $c, 2a, c$ values of $i$ respectively. So if the conjecture above holds, there has to be something special about polynomials that makes it work.

Let $p$ be an integer polynomial and $k$ be a natural number, both fixed. Is it the case that if

$$C(\alpha) = \sum_{i=1}^k e(\alpha \{p(i)/k\})$$

equals 0, then $\alpha$ is an integer? Here, $e(x) = e^{2 \pi i x}$ is the usual complex exponential and $\{x\}\in[0,1)$ is the fractional part of a real number $x$.

As strong as this conjecture may sound, it has been confirmed by all the examples that I have checked. For instance, if $p(i) = i$ and $k=4$, then

$$C(\alpha) = 1 + e(\alpha/4) + e(2\alpha/4) + e(3\alpha/4) = (1+e(\alpha/4))(1+e(\alpha/2))$$

is 0 if and only if $\alpha = 1,2,3 + 4\mathbb{Z}$.

Here is a potentially useful reformulation. Let $q$ be an integer polynomial for which $C(\alpha) = q(e(\alpha/k))$: in the example above, this is $$q(z) = 1 + z + z^2 + z^3 = (1+z)(1+z^2).$$ Then the conjecture is tantamount to showing that any root of $q$ on the unit circle is necessarily a $k$-th root of unity.

Note that the conjecture fails if we replace the set $\{p(1), \ldots, p(k)\}$ by arbitrary $b_1, \ldots, b_k\in\{0, \ldots, k-1\}$. For instance, if $(a,b,c)$ is a coprime Pythagorean triple and $$ q(z) = c \left(z + \frac{a}{c} + \frac{b}{c}i\right)\left(z + \frac{a}{c} - \frac{b}{c}i\right) = cz^2 + 2a z + c,$$ then this polynomial has roots on the unit circle that are not roots of unity, yet it corresponds to the sum $$ C(\alpha) = \sum_{i=1}^k e(\alpha b_i/k)$$ where $k=2(a+c)$ and $b_i$'s take values 0, 1 and 2 for $c, 2a, c$ values of $i$ respectively. So if the conjecture above holds, there has to be something special about polynomials that makes it work.