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For which polynomials $f(x)$ does there exist a nonconstant polynomial $g(x)$ such that $g\big(f(x)\big)$ is an even function?

If $f$ is already even, then $g$ can be the identity. If $f$ is odd, then $g(x)=x^2$ works. $f$ can be neither; for instance, if $f(x)=x+1$ then we can take $g(x)=(x-1)^2$.

However, not all polynomials work. A friend of mine pointed out that if $f(x)=x^2+x+1$, then $f(x)^n$ will always contain an $x^{2n-1}$ term that can't be canceled by lower powers, so there's no $g$ such that $g(x^2+x+1)$ is even.

A reasonable conjecture would be that such a $g$ exists iff $f$ is odd, even, or differs by an odd or even function by a constant. Is this true?

Akiva Weinberger
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  • Side note: WLOG $f(0)=0$ as is alluded to in the last paragraph. In that case, I believe there's always a nontrivial power series $g(x)$ such that $g(f(x))$ is even (one can construct its coefficients recursively). – Greg Martin Feb 07 '24 at 23:26
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    Can anyone show that there is no such $g(x)$ when $f(x)=x^4+x$? – Greg Martin Feb 07 '24 at 23:26
  • If $f(x) = x^{m+1}+ax^m+\cdots$, $a\neq 0$ and $m > 0$ then same reasoning as in the post implies there is no such $g$. – Jakobian Feb 07 '24 at 23:36

1 Answers1

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Write $f(x) = a_m x^m + \cdots + a_0$ and $g(x) = b_n x^n + \cdots + b_0$, where $a_m, b_m \neq 0$, and suppose that $g \circ f$ is even. Let $h_k(x) = f(x)^{k} - f(-x)^k$. Then

$$\sum_{k \leq n} b_k h_k(x) = 0,$$

implying that $\deg h_n = \deg \, (b_0h_0 + \cdots + b_{n-1}h_{n-1}) \leq \max_{k < n}(\deg h_k)$. The degree of $h_k$ can be found in three different cases:

  • if $f$ is even, then $\deg h_k = -\infty$.
  • if $f$ is not even, then:
    • if $m(k-1)$ is even, then $\deg h_k = m(k-1) + p$, where $p$ is the highest odd index with nonzero coefficient in $f$.
    • if $m(k-1)$ is odd, then $\deg h_k = m(k-1) + q$, where $q$ is the highest even index with nonzero coefficient in $f$.

Hence, if $f$ is not even, we get two potential inequalities:

  • if $m$ is even, then $m(n-1) + p \leq m(n-2) + p$, which is impossible.
  • if $m$ is odd and $n$ is even, then $m(n-1) + q \leq m(n-2) +p$, which implies that $m \leq p - q$, which is only possible if $p = m$ and $q = 0$. In other words, the only even degree term in $f$ is the constant term.

In conclusion, we must either have that $f$ is even, or that $f - a_0$ is odd.