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This question already has an answer here, but I'm looking for a solution that doesn't use field extensions. It's relatively easy to find a polynomial without zeros for every $n$ but as far as I know this doesn't imply that the polynomial is irreducible (maybe in a finite field it does?). Thank you for your answers.

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Let $F$ be a finite field of characteristic $p$ and let $n\in\Bbb{N}$. Let $k$ be an integer coprime to $p$. Then the polynomial $x^k+1$ is separable over $F$, so it is a product of distinct irreducible factors. For any degree $d$ there are only finitely many (irreducible) polynomials of degree $d$ with coefficients in $F$, so for sufficiently large values of $k$ it must have an irreducible factor of degree $n$.

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Hint use Euclid's idea: consider $1+p_1\cdots p_k$ where the $p_i$ are all irreducibles of degree $< n$

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  • $\begingroup$ If two polynomials have different values for every element in the field, does this imply they can't be the same polynomial? If so the answer follows quickly. $\endgroup$ – InsertNameHere Aug 15 at 15:05
  • $\begingroup$ OK, never mind, don't have an answer. $\endgroup$ – InsertNameHere Aug 15 at 15:33
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    $\begingroup$ @InsertNameHere It's not clear how your question is related to the above answer. Do you understand the answer? There are many prior answers that discuss formal vs. functional polynomials, e.g. here. $\endgroup$ – Bill Dubuque Aug 15 at 16:28

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