# Does there exist an even polynomial that has a term with an odd power?

Does there exist an even polynomial (i.e. $f(-x)=f(x)$ for all $x$) that has at least one odd power of $x$?

• In other words, is every even polynomial of the form $g(x^2)$ ? – lhf Apr 24 '14 at 18:10
• @lhf yes, that's right. I assume the answer is yes but I'm not sure what a rigorous proof would be.. – Superbus Apr 24 '14 at 18:11

## 6 Answers

Hint: You need $f(x) - f(-x)$ to be the zero polynomial.

• Why is the hint-only answer voted 16, while the proof answer is voted 5? – ignis Apr 25 '14 at 16:02
• Because it's much more valuable to look for the answer by yourself. – mak Apr 25 '14 at 17:00

The answer depends on which field you are working in. For example, over a field of characteristic $2$, every polynomial is even. Also, over a finite field $F_q$, we have that the polynomial $x^q - x$ is even (it vanishes for all $x\in F_q$) but its $x$ term has a non-zero coefficient.

However, over an infinite field of characteristic $\neq 2$, every even polynomial is a polynomial in $x^2$. This follows since $f(x) - f(-x) = 2g(x)$, where $g(x)$ consists of the odd powers of $x$ in $f$ and by assumption $2g(x)$ vanishes for all, and hence infinitely many, elements $x$. Thus $2g(x) = 0$ which implies $g(x) = 0$ (char $\neq 2$).

• Your comment about finite fields seems to be based on the identification of formal polynomials with polynomial functions. For the reason you describe, this is problematic in finite fields. – Ryan Reich Apr 25 '14 at 2:28
• I think the most natural definition of an even polynomial $P\in K[X]$ is one that does not change when $-X$ is substituted for$~X$. Then indeed in characteristic$~2$ all polynomials are even (the substitution is a no-op), but in other characteristics the even polynomials are exactly the ones without terms in odd degree, regardless of whether $K$ is finite or infinite. In particular $X^q-X$ is not even over $\Bbb F_q$ for an odd prime power$~q$. – Marc van Leeuwen Apr 25 '14 at 5:04

No. If $p(x)=a_nx^n+\cdots +a_1x+a_0$ then $$p(-x)=p(x)$$ $$a_0-a_1x+\cdots +(-1)^na_nx^n=a_0+a_1x+\cdots a_nx^n$$ Polynomials are identically equal exactly when their coefficients are equal, so $$a_1=-a_1\implies a_1=0,...$$ all coefficients of odd powers are $0$.

Alternatively, use induction. If $p$ is even, then $$-p'(-x)=p'(x)$$ so $p'$ is odd, and vice versa. Prove that constant and linear polynomials which are even and odd have only even and odd (respectively) terms, and you're done.

Split $\,f(x)\,$ into its even part $\,f_0(x) = \sum f_{2k} x^{2k}$ and odd part $\,f_1(x) = \sum f_{2k+1} x^{2k+1}$

Thus $\,f(x) = f_0(x)+ f_1(x)\$ where $\ f_0(-x) = f_0(x),\,$ and $\ f_1(-x) = -f_1(x)$

So $\ \ f(-x) = f_0(x)-f_1(x)\$

So $\ \ f(-x) = f(x)\!\iff\! 2 f_1(x) = 0\!\iff\! f_1(x) = 0\$ (assuming $\,2\,$ is cancellable).

Alternate answer. The sum of two even functions is even. So if $p$ is an even polynomial, add scalar multiples of even powers of $x$ to eliminate the even powers in $p(x)$. You'll be left with a supposedly even polynomial that has only odd powers of $x$. But that makes what you have an odd function. So you have a function that is even and odd, and it's easy to prove such a function is the zero function. Therefore there were no odd terms left over after cancelling the even terms.

Among smooth functions, the derivative of an even function is odd, and vice versa (and this is quite easy to prove). So if $p$ were even and there were an odd powered term, say $c_kx^k$, in a simplified expression for $p(x)$, then $\frac{d^k}{dx^k}p(x)$ would be an odd function (since $k$ is odd). But odd functions have $f(0)=0$, and this supposedly odd function would have a nonzero constant term $c_kk!$. So it's not possible.