Reciprocal divisibility of equally valued polynomials over a field

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Let $\mathbb {F}$ be a field.

Let $p(x),q(x) \in \mathbb{F}[x] s.t. \forall a \in \mathbb{F},p(a)=q(a) $.

Is it true, in general, that $p(x)|q(x) \vee q(x)|p(x)$?

It seems plausible for me that this holds even in finite fields, as I expect the result of the division to be 1, but I've tried some examples, dividing via Ruffini's method, and this seems not to hold, but I can't figure out why.

Thank you.

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In a finite field $F$ of $q$ elements, the polynomial $f(x)=x^q-x$ has the property that $f(a)=0$ for all elements of $F$. Therefore so do $xf(x)$ and $(x+1)f(x)$.