The \(n\)th degree polynomial P\((x)\) is said to be reflexive if:
P\((x)\) is of the form \(x^n - a_1x^{n-1} + a_2x^{n-2} - \cdots + (-1)^na_n\) where \(n \geq 1\);
\(a_1, a_2, \ldots, a_n\) are real;
the \(n\) (not necessarily distinct) roots of the equation P\((x) = 0\) are \(a_1, a_2, \ldots, a_n\).
Find all reflexive polynomials of degree less than or equal to 3.
For a reflexive polynomial with \(n > 3\), show that
$$2a_2 = -a_2^2 - a_3^2 - \cdots - a_n^2.$$
Deduce that, if all the coefficients of a reflexive polynomial of degree \(n\) are integers and \(a_n \neq 0\), then \(n \leq 3\).
Determine all reflexive polynomials with integer coefficients.
Solution:
Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\).
Suppose \(n = 2\), then we want
\begin{align*}
&& x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\
&&&= x^2-(a_1+a_2)x+a_1a_2 \\
\Rightarrow && a_2 &= 0 \\
\end{align*}
So all polynomials of the form \(x^2-a_1x\) work and no others.
Suppose \(n = 3\) then we want
\begin{align*}
&& x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\
&&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\
\Rightarrow && a_2+a_3 &= 0 \\
&& a_2a_3 &= a_2 \\
\Rightarrow && -a_2^2 &= a_2 \\
\Rightarrow && a_2 &= 0, -1 \\
&& -a_1a_2^2 &= -a_2 \\
\Rightarrow && a_2 &= 0, a_2 = 1/a_1
\end{align*}
So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
Suppose \(n > 3\) then
\begin{align*}
&& \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\
&& &= a_1^2 - 2a_2 \\
\Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\
&&&= -a_2^2 - a_3^2 - \cdots - a_n^2
\end{align*}
So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)