Problem - 2840
Let $n$ be an even positive integer, and let $p(x)$ be an $n$-degree polynomial such that $p(-k) = p(k)$ for $k = 1, 2, \dots , n$. Prove that there is a polynomial $q(x)$ such that $p(x) = q(x^2)$.
Let $p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$. Then the $(n-1)$-degree polynomial $$p(x) - p(-x) = 2(a_1x+a_3x^3+\cdots+a_{n-1}x^{n-1})$$ equals zero at $n$ different points. Therefore, it must be identically zero, i.e., $a_1 = a_3 = \cdots = a_{n-1} = 0$. It follows that $$p(x) = a_0+a_2x^2+a_4x^4 +\cdots+a_nx^n$$ and $$q(x) = a_0 + a_2x + a_4x^2 +\cdots + a_nx^{\frac{n}{2}}$$ meet the requirement.