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Is $x^3-9$ irreducible over the integers mod 31?

I know that one method is to check whether it has any roots or not, but that turns out to be very tedious in this case.

So, is there any other simpler method to find it out?

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1 Answer

The multiplicative group of the finite field $F_{31}=\mathbf{Z}/31\mathbf{Z}$ is cyclic of order 30. Therefore the non-zero cubes form a cyclic subgroup of order 10. That subgroup consists of all the elements of order dividing ten, so all you need to do is to check, whether $9^{10}\equiv 1\pmod{31}$ or not. Leaving that to you.

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