Even multiplicities at fixed points of reciprocal root pairing ($P(x)=x^{2m}P(1/x)⟹{\mathrm{ord}}_{1}P,{\mathrm{ord}}_{-1}P\in 2\mathbb{Z}$)

Let $P$ be a nonzero polynomial of even degree $2m$ over a field of characteristic different from two, satisfying the displayed reciprocal identity. If $P(x)=(x-1{)}^{r}A(x)$ with $A(1)\ne 0$, cancellation gives $A(x)=(-1{)}^r{x}^{2m-r}A(1/x)$. Substitution of $x=1$ forces $r$ even. If $P(x)=(x+1{)}^{s}B(x)$ with $B(-1)\ne 0$, 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.