The Modulus of all the roots of a Polynomial are equal to $1$ 
Suppose the real number $\lambda \in (0,1)$, and let $n$ be a positive integer. Prove that all 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 equal to $1.$

The Putnam problem 2014 B4 is similar:  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.
 A: The following solution is taken from The Modulus of a Polynomial are the Same on AoPS. For other solutions see A family of polynomials whose zeros all lie on the unit circle on Math Overflow.
The idea is to derive a recursion formula for
$$
f_n(z) = \sum_{k=0}^{n} \binom{n}{k}\lambda^{k(n-k)} z^k 
$$
and prove the statement by induction. $\lambda$ is a fixed real number in the interval $(0, 1)$.
$f_0(z) = 1$ and $f_1(z) = 1+z$ surely have only roots of modulus one.
Now assume that $n \ge 1$ and all roots of $f_n$ have modulus one. Using the recurrence formula for binomial coefficients
$$
 \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}
$$
one gets
$$
 f_{n+1}(z) = f_n(\lambda z) +  \lambda^n z f_n(\frac{z}{\lambda}) \, .
$$
Now let $z^*$ be a zero of $f_{n+1}$. We have to show that $|z^*| = 1$.
From the above recurrence formula we get that
$$
 f_n(\lambda z^*) = -  \lambda^n z^* f_n(\frac{z^*}{\lambda})
$$
Note that $f_n(\frac{z^*}{\lambda})$ cannot be zero: Otherwise both $\lambda z^*$ and $\frac{z^*}{\lambda}$ would have modulus one, which is not possible.
Denoting the roots of $f_n$ with $z_1, \ldots z_n$ we have $f_n(z) = (z-z_1) \cdots (z-z_n)$ so that
$$
 z^* = - \frac{f_n(\lambda z^*)}{\lambda^n f_n(\frac{z^*}{\lambda})}
= - \prod_{k=1}^n \frac{\lambda z^* - z_k}{z^* - \lambda z_k}
$$
and therefore
$$ \tag{*}
 |z^*| = \prod_{k=1}^n \left |\frac{\lambda z^* - z_k}{z^* - \lambda z_k} \right | \, .
$$
An elementary calculation (using $|z_k|=1$ for all $k$) shows that
$$
|\lambda z^* - z_k|^2 - |z^* - \lambda z_k|^2 = (1 - \lambda^2)(1 - |z^*|^2) \, .
$$
It follows that if $|z^*| < 1$ then all factors on the right-hand side of $(*)$ are larger than one, and if $|z^*| > 1$ then all factors on the right-hand side of $(*)$ are less than one. So both cases lead to a contradiction, and the only possibility is that $|z^*| = 1$. This concludes the proof.
A: Write the above polynomial in the form:     $a+bx+cx^2$... and observe that if we put $f(x)=0$ it is same as putting $x^k f(1/x)=0$ {property of binomial coefficients and the power of lambda} conclude that if $x$ is a root  then $1/x$ is also has the same modulus. Hence $|x|=1$ 
