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Let $k$ be a field of characteristic zero, and let $f=f(x) \in k[x]$ be a polynomial which satisfies $f(x)+f(-x+u)=0$, where $u \in k$.

My question: Is there an easy way to find the exact form of such $f$?

Please notice that the special case in which $u=0$ is easy: If $f(x)+f(-x)=0$, then $f(x)= -f(-x)$, so $f$ is odd.

However, for $u \in k^*$ the situation is slightly more complicated; such $f$ need not be odd, for example: $f(x)=ax-\frac{au}{2}$, with $a \in k^*$, is not odd (and not even).

Also notice that such $f$ satisfies $f(\frac{u}{2})=0$, by taking $x=\frac{u}{2}$ in $f(x)+f(-x+u)=0$ (the characteristic of $k$ is zero by assumption), namely $\frac{u}{2}$ is a root of $f$, so $x-\frac{u}{2}$ divides $f$.

Similar conditions are also satisfied, for example, $f(u)=-f(0)$, $f(\frac{u}{3})=-f(\frac{2u}{3})$, etc.

(Perhaps considering its derivatives may help?).

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    If the characteristic of $k$ is not 2, then $g(x) = f(x + u/2)$ is odd, and conversely if $g$ is odd then $f(x) = g(x - u/2)$ is of the form you describe. Is that what you're looking for? – 2017-02-26
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    I am not sure yet.. But your idea is very nice and may be good enough as an answer. Please, do you think that there is no easy way to describe the coefficients of $f$? – 2017-02-26
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    @MeesdeVries How are you defining odd in positive characteristic? – 2017-02-26
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    @MichaelBurr, as $\forall x(f(-x) = -f(x))$; is there anything wrong with that? – 2017-02-26
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    @MeesdeVries I was thinking an odd value and not an odd function; slow day for me... – 2017-02-26

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