Putnam 2023, Problem A-2

Let \( n \) be an even positive integer. Let \( p \) be a monic, real polynomial of degree \( 2n \); that is to say, \( p(x) = x^{2n} + a_{2n-1}x^{2n-1} + \cdots + a_1x + a_0 \) for some real coefficients \( a_0, \ldots, a_{2n-1} \). Suppose that \( p(1/k) = k^2 \) for all integers \( k \) such that \( 1 \leq |k| \leq n \). Find all other real numbers \( x \) for which \( p(1/x) = x^2 \).

Solution:
The only other real numbers with this property are $\pm 1/n!$. (Note that these are indeed other values than $\pm 1, \ldots, \pm n$ because $n > 1$.)

Define the polynomial $q(x) = x^{2n+2} - x^{2n}p(1/x) = x^{2n+2} - (a_0x^{2n} + \cdots + a_{2n-1}x + 1)$. The statement that $p(1/x) = x^2$ is equivalent (for $x \neq 0$) to the statement that $x$ is a root of $q(x)$. Thus we know that $\pm 1, \pm 2, \ldots, \pm n$ are roots of $q(x)$, and we can write
$$q(x) = (x^2 + ax + b)(x^2 - 1)(x^2 - 4) \cdots (x^2 - n^2)$$
for some monic quadratic polynomial $x^2 + ax + b$. Equating the coefficients of $x^{2n+1}$ and $x^0$ on both sides gives $0 = a$ and $-1 = (-1)^n(n!)^2b$, respectively. Since $n$ is even, we have $x^2 + ax + b = x^2 - (n!)^{-2}$. We conclude that there are precisely two other real numbers $x$ such that $p(1/x) = x^2$, and they are $\pm 1/n!$.