TITLE: A family of polynomials whose zeros all lie on the unit circle
QUESTION [35 upvotes]: I had posted the following problem on stack exchange  before.

Suppose $\lambda$ is a real number in $\left( 0,1\right)$, and let $n$ be a positive integer. Prove that all the roots of the polynomial $$
f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}
$$ have modulus $1$.

I do not seem to know how to do it, to show if the roots of the polynomial have modulus one.  Putnam 2014 B4 Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers.
This problem  is very similar to the Putnam problem.

REPLY [22 votes]: This follows from Julienne's answer. Fix $\lambda \in (0,1)$ and let
$$ f_n(x) = \sum_{k=0}^n {n \choose k} \lambda^{k(n-k)}x
^k. $$
Each $f_n$ satisfies condition (a) for $\mu = 1$. As for (b), we can compute the derivative of $f_n$ to get
$$f_n ' (x) = n \lambda^{n-1} f_{n-1} \left(\frac{x}{\lambda}\right).$$
Now use induction on $n$: for $n=1$ we have $f_1(x) = 1+x$, so $x=-1$ is its only root. If $f_{n-1}$ has all roots on the unit circle, then all roots $x_0$ of $f_n'$ satisfy
$$n\lambda^{n-1}f_{n-1}\left( \frac{x_0}{\lambda}\right)=0$$
and so $x_0=\lambda z_0$ for some $z_0$ on the unit circle, by the induction hypothesis. Hence, as $\lambda \in (0,1)$, $x_0$ lies inside the unit circle. By induction on $n$, (b) holds for all $f_n$ and so the result follows from Cohn. 
(Sorry for not leaving this as a comment to Julienne's answer, I do not have enough reputation to comment.)
Edit: corrected formula for $f_n'$ thanks to Ian Agol's correction.