Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z)-1$ have absolute value 1 .
The answer is $P(x)$ should be a polynomial of the form $P(x)=\lambda x^{n}-\mu$ where $|\lambda|=|\mu|$ and $\operatorname{Re} \mu=-\frac{1}{2}$. One may check these all work; let's prove they are the only solutions.
First approach (Evan Chen). We introduce the following notations:
$$
\begin{aligned}
P(x) & =c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{1} x+c_{0} \\
& =c_{n}\left(x+\alpha_{1}\right) \ldots\left(x+\alpha_{n}\right) \\
P(x)-1 & =c_{n}\left(x+\beta_{1}\right) \ldots\left(x+\beta_{n}\right)
\end{aligned}
$$

By taking conjugates,
$$
\begin{aligned}
\left(x+\alpha_{1}\right) \cdots\left(x+\alpha_{n}\right) & =\left(x+\beta_{1}\right) \cdots\left(x+\beta_{n}\right)+c_{n}^{-1} \\
\Longrightarrow\left(x+\frac{1}{\alpha_{1}}\right) \cdots\left(x+\frac{1}{\alpha_{n}}\right) & =\left(x+\frac{1}{\beta_{1}}\right) \cdots\left(x+\frac{1}{\beta_{n}}\right)+\left(\overline{c_{n}}\right)^{-1}
\end{aligned}
$$

The equation $(\boldsymbol{\Phi})$ is the main player:
Claim - We have $c_{k}=0$ for all $k=1, \ldots, n-1$.

Proof. By comparing coefficients of $x^{k}$ in ( $\boldsymbol{\Lambda}$ ) we obtain
$$
\frac{c_{n-k}}{\prod_{i} \alpha_{i}}=\frac{c_{n-k}}{\prod_{i} \beta_{i}}
$$
but $\prod_{i} \alpha_{i}-\prod_{i} \beta_{i}=\frac{1}{c_{n}} \neq 0$. Hence $c_{k}=0$.
It follows that $P(x)$ must be of the form $P(x)=\lambda x^{n}-\mu$, so that $P(x)=\lambda x^{n}-(\mu+1)$. This requires $|\mu|=|\mu+1|=|\lambda|$ which is equivalent to the stated part.

Second approach (from the author). We let $A=P$ and $B=P-1$ to make the notation more symmetric. We will as before show that $A$ and $B$ have all coefficients equal to zero other than the leading and constant coefficient; the finish is the same.

First, we rule out double roots.
Claim - Neither $A$ nor $B$ have double roots.
Proof. Suppose that $b$ is a double root of $B$. By differentiating, we obtain $A^{\prime}=B^{\prime}$, so $A^{\prime}(b)=0$. However, by Gauss-Lucas, this forces $A(b)=0$, contradiction.
Let $\omega=e^{2 \pi i / n}$, let $a_{1}, \ldots, a_{n}$ be the roots of $A$, and let $b_{1}, \ldots, b_{n}$ be the roots of $B$. For each $k$, let $A_{k}$ and $B_{k}$ be the points in the complex plane corresponding to $a_{k}$ and $b_{k}$.
Claim (Main claim) — For any $i$ and $j, \frac{a_{i}}{a_{j}}$ is a power of $\omega$.

Proof. Note that
$$
\frac{a_{i}-b_{1}}{a_{j}-b_{1}} \cdots \frac{a_{i}-b_{n}}{a_{j}-b_{n}}=\frac{B\left(a_{i}\right)}{B\left(a_{j}\right)}=\frac{A\left(a_{i}\right)-1}{A\left(a_{j}\right)-1}=\frac{0-1}{0-1}=1 .
$$

Since the points $A_{i}, A_{j}, B_{k}$ all lie on the unit circle, interpreting the left-hand side geometrically gives
$$
\measuredangle A_{i} B_{1} A_{j}+\cdots+\measuredangle A_{i} B_{n} A_{j}=0 \Longrightarrow n \widehat{A_{i} A_{j}}=0,
$$
where angles are directed modulo $180^{\circ}$ and arcs are directed modulo $360^{\circ}$. This implies that $\frac{a_{i}}{a_{j}}$ is a power of $\omega$.
Now the finish is easy: since $a_{1}, \ldots, a_{n}$ are all different, they must be $a_{1} \omega^{0}, \ldots, a_{1} \omega^{n-1}$ in some order; this shows that $A$ is a multiple of $x^{n}-a_{1}^{n}$, as needed.