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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.

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    $\begingroup$ The polynomial has "symmetrical" coefficents and can therefore be written as $g(\frac{x+\frac{1}{x}}{2})$ where $g$ is a polynomial. Then the problem is reduced to showing that $g$ has all its roots real and inside $[-1,1]$. Here techniques as in the Putnam question mentioned in the OP would probably be useful. $\endgroup$ – Ewan Delanoy May 3 '18 at 11:38
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    $\begingroup$ Also posted on MO: mathoverflow.net/questions/299304/… $\endgroup$ – Wolfgang May 3 '18 at 12:19
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    $\begingroup$ @EwanDelanoy Well, $x+1$ has “symmetric” coefficients but how to write it as a polynomial of $\dfrac12\left(x+\dfrac1x\right)$? $\endgroup$ – Saad May 8 '18 at 9:00
  • $\begingroup$ @AlexFrancisco This is a classical exercise : by induction on $n$, one shows that $x^n+\frac{1}{x^n}$ is of the form $Q_n(x+\frac{1}{x})$ where $Q_n$ is a uniquely defined polynomial of degree $n$. The exact term is "palindromic", by the way. $\endgroup$ – Ewan Delanoy May 8 '18 at 10:09
  • $\begingroup$ @EwanDelanoy Isn't this true only for polynomials of even degree? $\endgroup$ – Servaes May 11 '18 at 14:26
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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 $|x|=1$

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