\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We look at different cases depending on the value of \(n\).
When \(n = 2\), \(P(x) = x^2 - a_1 x + a_2\) has root \(a_1, a_2\), and hence by Vieta’s Theorem, \[ a_1 a_2 = a_2, a_1 + a_2 = a_1. \]
This means \(a_2 = 0\) and \(a_1\) can take any real value, and hence \[ P(x) = x^2 - a_1 x \] is reflective for \(a_1 \in \RR \).
When \(n = 3\), \(P(x) = x^3 - a_1 x^2 + a_2 x - a_3\) has root \(a_1, a_2, a_3\), and hence by Vieta’s Theorem, \[ \left \{ \begin {aligned} a_1 a_2 a_3 & = a_3, \\ a_1 a_2 + a_1 a_3 + a_2 a_3 & = a_2, \\ a_1 + a_2 + a_3 & = a_1. \end {aligned} \right . \]
The final equation implies that \(a_2 + a_3 = 0\), and hence with the second equation gives that \(a_2 a_3 = a_2\), which means either \(a_2 = a_3 = 0\), or \(a_2 = -1, a_3 = 1\).
When \(a_2 = a_3 = 0\), \(a_1\) can take any real value, and when \(a_2 = -1, a_3 = 1\), we must have \(a_1 = -1\).
So the degree \(3\) reflective polynomials are \[ P(x) = x^3 - a_1 x^2 \] for all \(a_1 \in \RR \), and \[ P(x) = x^3 + x^2 - x - 1. \]
By Vieta’s Theorem, we have \[ \sum _{i = 1}^{n} a_i = a_1, \] and hence \[ \sum _{i = 2}^{n} a_i = 0. \]
Squaring both sides gives \begin {align*} 0 & = \left (\sum _{i = 2}^{n} a_i\right )^2 \\ & = \sum _{i = 2}^{n} a_i^2 + 2 \sum _{i = 2}^{n - 1} \sum _{j = i + 1}^{n} a_i a_j. \end {align*}
By Vieta’s Theorem, we also have \[ \sum _{i = 1}^{n - 1} \sum _{j = i + 1}^{n} a_i a_j = a_2, \] and notice that \begin {align*} 2 a_2 & = 2 \sum _{i = 1}^{n - 1} \sum _{j = i + 1}^{n} a_i a_j \\ & = 2 \sum _{j = 2}^{n} a_1 a_j + 2 \sum _{i = 2}^{n - 1} \sum _{j = i + 1}^{n} a_i a_j \\ & = 2 a_1 \sum _{i = 2}^{n} a_i + \left (-\sum _{i = 2}^{n} a_i^2\right ) \\ & = 2 a_1 \cdot 0 - \sum _{i = 2}^{n} a_i^2 \\ & = -\sum _{i = 2}^{n} a_i^2, \end {align*}
as desired.
For the final part, assume B.W.O.C. that \(n > 3\). By rearrangement, we have \begin {align*} a_2^2 + 2 a_2 + 1 = 1 - \sum _{i = 3}^{n} a_i^2, \end {align*}
and the left-hand side is \((a_2 + 1)^2\) which is always non-negative. Hence, \[ \sum _{i = 3}^{n} a_i^2 \leq 1. \]
Since \(a_i\) are all integers, precisely one of the \(a_i\)s for \(3 \leq i \leq n\) is \(\pm 1\), and all the rest are \(0\). Since \(a_n \neq 0\), we conclude that \(a_n = \pm 1\), and \(a_3 = \cdots = a_{n - 1} = 0\).
But notice from Vieta’s Theorem that \[ a_n = \prod _{i = 1}^{n} a_i = 0 \] since \(a_3\) must be \(0\), which leads to a contradiction.
Hence, we must have \(n \leq 3\).
The reflective polynomials for \(n \leq 3\) are
For \(n > 3\), we must have \(a_n = 0\), and hence \begin {align*} P(x) & = x^n - a_1 x^{n - 1} + a_2 x^{n - 2} - \cdots + (-1)^{n - 1} a_{n - 1} x \\ & = x \left (x^{n - 1} - a_2 x^{n - 2} + a_2 x^{n - 3} - \cdots + (-1)^{n - 1} a_{n - 1}\right ). \end {align*}
Let \[ Q(x) = x^{n - 1} - a_2 x^{n - 2} + a_2 x^{n - 3} - \cdots + (-1)^{n - 1} a_{n - 1} \]
If \(P(x)\) is reflective, then the roots to \(P(x)\) are \(a_1, a_2, \ldots , a_{n - 1}, 0\), and hence the roots to \(Q(x)\) are \(a_1, a_2, \ldots , a_{n - 1}\), which shows that \(Q(x)\) is reflective as well.
This means that an integer-coefficient reflective polynomial with degree \(n > 3\) must be \(x\) multiplied by another integer-coefficient reflective polynomial, and repeating this process, we can conclude it must be some power of \(x\) multiplied by some integer-coefficient reflective polynomial with degree \(n \leq 3\).
Hence, all integer-coefficient reflective polynomials are