Let $p$ be a prime. Consider a sequence $S$ of $p$ binary numbers $x_n \in \{ 0, 1 \}$, i.e. $S = \{x_1, x_2, \cdots, x_p\}$, where the number of zeroes in $S$ is neither $0$ nor $p$. Then the conjecture is: for all such sequences, all discrete fourier transformation (DFT) components $\hat S_k$ of $S$ are nonzero. That is, show that, $\forall k = 1\cdots p$, one obtains $\hat S_k = \sum_{n=1}^p x_n \exp (2 \pi i \frac{ n k }{p}) \ne 0$.
A way to the solution might be the visualization that a vectorial sum of $p$ evenly distributed spokes in a wheel (the unit circle) is performed, where only those spokes with $x_n = 1$ are existent, and where every $k$th spoke (modulo $p)$ is counted, and the vector sum will not disappear.
It is clear that the conjecture only holds if $p$ is prime. Since if it weren't, write $p= q \cdot r$, and consider a sequence $S^q$ where every $q$th entry is $x_n = 1$ and all others are $0$. The DFT components $\hat S^q_k$ of $S^q$ will then be $$\hat S^q_k = \sum_{n=1}^p x_n \exp (2 \pi i\frac{ n k }{q r}) = \sum_{n=1}^r \exp (2 \pi i \frac{ n q k }{q r}) = \sum_{n=1}^r \exp (2 \pi i \frac{ n k }{r}) = 0.$$
This DFT $\hat S_k$ equals $f(\zeta_k)$, where $f(t) = x_p + \sum_{n=1}^{p-1} x_n t^n$ is a polynomial with integer coefficients and $\zeta_k = \exp(2\pi i k/p)$ is a primitive $p$th root of unity (since $p$ is prime). The only way $f(\zeta_k)$ could equal $0$ is if $f(t)$ is a multiple of the minimal polynomial of $\zeta_k$, which is $1+t+t^2+\cdots+t^{p-1}$. There are only two cases where this happens, namely $x_1=\cdots=x_p=0$ and $x_1=\cdots=x_p=1$; this finishes the proof.
(Proof that $\zeta_k$ is a primitive $p$th root of unity: it is a $p$th root of unity since $\zeta_k^p = \exp(2\pi i k/p)^p = \exp(2\pi i k p/p) = \exp(2\pi i k) = 1$, but for any $1\le m\le p-1$ we have $\zeta_k^m = \exp(2\pi i k/p)^m = \exp(2\pi i k m/p) \ne 1$ since $km$ is not a multiple of $p$.)