Let $h(X) := \frac{f(X)}{g(X)} ∈ ℚ(X) \setminus ℚ$ be such that $h(-X) = h(X)$. Does that imply that all odd coefficients of $f$ and $g$ are $0$? Or can this be accomplished in other ways?
2026-03-29 11:44:28.1774784668
$\frac{f(X)}{g(X)} = \frac{f(-X)}{g(-X)} ∈ ℚ(X)$
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Lacking any further constraints on $f$ and $g$, no. Consider $f(X) = g(X) = X$.
As to other ways, why not $h(X) \in \Bbb{Q}(X^2)$?