Conjecture: If a polynomial in $\mathbb{C[x]}$ take only real values on the unit circle, then its identically equal to a real constant.
It's quite easy to see that the statement can be reformulated as:
Let $r_1,...,r_n$ and $\phi_1,...\phi_n$ be real numbers. If $S(\alpha)=\sum\limits_{k=1}^{n}r_k\sin(\phi_k+k\alpha)$ is constant for all $\alpha \in \mathbb{R}$, then $r_1=r_2=...=r_n=0$.
This seems intuitively true, because its essentially saying that we can't have a sum of harmonics which sum to zero.
Can I show the reformulation via simples means without resorting to Fourier series?
The functions $e_k:\mathbb R\to\mathbb C$ defined by $e_k(\theta)=e^{ik\theta}$ are linearly independent for any set of distinct $k$, in particular for $k\in\mathbb Z$. One way to see this is because eigenvectors for distinct eigenvalues of a linear transformation on a vector space are linearly independent, applied on the complex vector space $V$ of infinitely differentiable functions from $\mathbb R$ to $\mathbb C$ with the linear transformation being differentiation, $D:V\to V$, $Df = f'$. Note that $De_k=ike_k$, so $e_k$ is an eigenvector for $D$ with eigenvalue $ik$.
Suppose $f\in \mathbb C[x]$, so $f(x)=\sum\limits_{k=0}^n a_k x^k$. On the unit circle, $f(e^{i\theta})=\sum\limits_{k=0}^n a_k e^{ik\theta}$. If this quantity is always real, then $a_0-\overline{a_0}+\sum\limits_{k=1}^{n}a_k e^{ik\theta}+\sum\limits_{k=-n}^{-1}-\overline{a_{-k}}e^{ik\theta}=0$ for all $\theta\in\mathbb R$. In other words, $$(a_0-\overline{a_0})e_0+\sum\limits_{k=1}^{n}a_k e_k+\sum\limits_{k=-n}^{-1}-\overline{a_{-k}}e_k=0.$$
Because $\{e_k: k\in\{-n,-n+1,\ldots,n-1,n\}\}$ is linearly independent, each coefficient must be zero, so $a_0$ is real and $a_k=0$ for $k>0$.