= Even multiplicities at fixed points of reciprocal root pairing
{title2=$P(x)=x^{2m}P(1/x)\Longrightarrow\operatorname{ord}_{1}P,\operatorname{ord}_{-1}P\in2\mathbb Z$}
Let $P$ be a nonzero <polynomial> of even degree $2m$ over a <field> of <characteristic> different from two, satisfying the displayed reciprocal identity. If $P(x)=(x-1)^rA(x)$ with $A(1)\ne0$, cancellation gives $A(x)=(-1)^r x^{2m-r}A(1/x)$. Substitution of $x=1$ forces $r$ even. If $P(x)=(x+1)^sB(x)$ with $B(-1)\ne0$, cancellation gives $B(x)=x^{2m-s}B(1/x)$; substitution of $x=-1$ forces $s$ even. Zero multiplicities are included. Pairing nonfixed reciprocal roots alone cannot prove this parity, because $1$ and $-1$ are their own inverses.
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