Nineteenth IMAR Mathematical Competition

Leaving aside the trivial cases $n=0$ and $n=1$, assume $n\ge 2$. The conclusion follows from the fact that the polynomial under consideration,

$$f=\sum _{k=0}^n{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2(X+1{)}^{2k}(X-1{)}^{2(n-k)},$$ is expressible as a sum of $n+1$ polynomials of the form $(b_k{X}^2+c_k{)}^n$, where the $b_k$ and the $c_k$ are all non-negative real numbers and at least $n-1$ products $b_k{c}_k$ are positive; this latter then implies that the even powers of $X$ all occur in the expansion with a positive coefficient. In particular, the fact that the odd rank coefficients in the standard expansion all vanish comes for free.

Let $S$ be the set of $(n+1)$-st roots of unity and recall that, if $k$ and $\ell$ are integers in the range $0$ through $n$, then the sum ${\sum }_{\omega \in S}{\omega }^k{\bar{\omega }}^{\ell }$ vanishes, unless $k=\ell$ in which case it is equal to $n+1$.

$$\begin{array}{rl}f& =\frac{1}{n+1}\sum _{k=0}^n\sum _{\ell =0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)(X+1{)}^{k+\ell }(X-1{)}^{2n-k-\ell }\sum _{\omega \in S}{\omega }^k{\bar{\omega }}^{\ell }\\ & =\frac{1}{n+1}\sum _{\omega \in S}\left(\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\omega }^k(X+1{)}^k(X-1{)}^{n-k}\right)\\ & \cdot \left(\sum _{\ell =0}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right){\bar{\omega }}^{\ell }(X+1{)}^{\ell }(X-1{)}^{n-\ell }\right)\\ & =\frac{1}{n+1}\sum _{\omega \in S}(\omega (X+1)+X-1{)}^n(\bar{\omega }(X+1)+X-1{)}^n\\ & =\frac{1}{n+1}\sum _{\omega \in S}{\left((2+\omega +\bar{\omega })X^2+2-\omega -\bar{\omega }\right)}^n.\end{array}$$

Finally, since each $\omega +\bar{\omega }$ is a real number whose absolute value does not exceed $2$ and $\omega +\bar{\omega }<2$ for at least $n-1$ roots $\omega \ne \pm 1$, each summand above is a polynomial of the desired form. This completes the proof.