Uniqueness of the squared polynomial in the Fejer-Riesz Lemma I know that if $f$ is a nonnegative trigonometric polynomail of degreen $n$, it is the square of a trigonometric polynomial $p$ that belongs to the span of $1$, $e^{ix}$, ..., $e^{inx}$. That is, $f=\mid p \mid ^2$.
I also know that $p$ can be defined as a polynomial on $\mathbb{C}$ that is nonzero in the open unit disc $\mathbb{D}$.
But then I want to show that such $p$ is unique up to the multiplicative constant of absolute value $1$. I though it would be easy to show, but turns out to be far trickier.... 
Could anyone help me?
 A: The proof below follows Conrad's idea and the discussion with OP. I am here just writing out the details.

Claim. If the polynomial is chosen to have NO root in $\mathbb{D}=\{z:|z|<1\}$, then it is unique up to a constant of modulus one. 

Firstly, I need to note that if a polynomial $P$ is chosen to have NO root in $\mathbb{D}$, then it has no root in $\overline{D}^{c}=\{z:|z|>1\}$ either. Otherwise, if $\alpha\in \overline{D}^{c}$ was a root for $P$, then $P(\alpha)=0$, so $f(\alpha)=|P(\alpha)|^{2}=0$. Also, $\alpha\in\overline{\mathbb{D}}^{c}\implies\alpha\neq 0$, so by the proof of the Riesz-Fejer Theorem, we know that $f(\overline{\alpha}^{-1})=0$. Then $|P(\overline{\alpha}^{-1})|^{2}=|f(\overline{\alpha}^{-1})|^{2}=0$. But $\overline{\alpha}^{-1}\in\mathbb{D}$, a contradiction. 
Therefore, all the roots of $P$ lie on $\mathbb{S}^{1}$. 
Suppose $P$ and $Q$ are two such polynomials and $|P|^{2}=|Q|^{2}$ on $\mathbb{S}^{1}$. Since they are polynomials, they must have only finitely many zeros. 
Also, they must have the same zeros, including multiplicities. Indeed, $|P|^{2}=|Q|^{2}$ on $\mathbb{S}^{1}$ implies that there is some polynomial with real coefficients, say $G(z)$, is given by $P(z)$ times its conjugate or $Q(z)$ times its conjugate, for $z\in\mathbb{S}^{1}$. Since $G$ is the same polynomial, $P$ and $Q$ must have same zeros counting multiplicities on $\mathbb{S}^{1}$. But all of their zeros lie on $\mathbb{S}^{1}$, so we are done. 
Now, since $Q$ has no zero inside $\mathbb{D}$, $P/Q$ is analytic in $\mathbb{D}$. But since they have the same zeros counting multiplicities, by the continuity over the finite many of them, we can extend analytically $P/Q$ on $\overline{\mathbb{D}}$ to a function with modulus $1$. 
Therefore, by the maximum modulus principle, we know that for all $z\in\mathbb{D}$, $$\Big|\dfrac{P(z)}{Q(z)}\Big|\leq\sup_{z\in\mathbb{D}}\Big|\dfrac{P(z)}{Q(z)}\Big|\leq\sup_{z\in\overline{\mathbb{D}}\setminus\mathbb{D}}\Big|\dfrac{P(z)}{Q(z)}\Big|=1.$$
Therefore, $|P(z)|\leq |Q(z)|$ for all $z\in\mathbb{D}$. 
Applying the similar argument to $Q/P$, we will have $|Q(z)|\leq|P(z)|$ for all $z\in\mathbb{D}$.  
Overall, we know that $|P(z)|=|Q(z)|$ for all $z\in\mathbb{D}$. 
This implies that the range of $P/Q$ mapping from $\mathbb{D}$ is confined in $\overline{\mathbb{D}}$, and thus $P/Q$ is closed. 
Thus, by the open mapping theorem, we know that $P/Q$ must be a constant in $\mathbb{D}$, namely, for all $z\in\mathbb{D}$, we have $$P(z)=cQ(z)\ \text{for some constant}\ c.$$
But we know that $|P/Q|=1$, and thus $|c|=1$. 
Now, recall the identity theorem that given functions $f$ and $g$ holomorphic on a domain $D$, if $f=g$ on some $S\subset D$, $S$ having a limit point, then $f=g$ on the whole $D$. 
To apply the theorem, we take $f=P$ and $g=cQ$, $D=\mathbb{C}$ and $S=\mathbb{D}$. Then $f$ and $g$ are just polynomial, so they are holomorphic in the whole $\mathbb{C}$. $\mathbb{C}$ is clearly a domain in itself. $\mathbb{D}$ is open in $\mathbb{C}$, so every point is a limit point. 
Since we have showed that $P(z)=cQ(z)$ for all $z\in\mathbb{D}$ with $c$ a constant of modulus $1$, then it follows immediately from the the identity theorem that for all $z\in\mathbb{C}$, we always have $$P(z)=cQ(z),\ \text{with}\ c\ \text{a constant of modulus}\ 1.$$
Edit 1:
I think this claim is not true if we assume the polynomial has NO root in $\overline{\mathbb{D}}^{c}$. Because in this way, we cannot say all the roots lying on $\mathbb{S}^{1}$. See the first paragraph of the proof, we need $\alpha\neq 0$, because we need to use the fact that: $f(z)=\overline{f(z^{-1})}$ for all $z\neq 0$. However, if argue by contradiction and suppose there is a root inside $\mathbb{D}$, then we cannot say $\alpha\neq 0$. 
I think this is why the uniqueness argument only happens for polynomial not having root in $\mathbb{D}$. 
