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

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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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.
consider G = Z/31Z ,the group of nonzero elements of Z/31Z under multiplication. This is a cyclic group of order 30. The map $\Phi: G \to G$ defined by $\Phi(x) = x^3$ is a homomorphism. So we need to know if 9 is contained in the image of Φ. Note that the kernel of $\Phi$ is the unique subgroup of order 3 in G. So the image has order 30/3 = 10. So we need to know if 9 is in the subgroup of order 10. Now computing the order of 9 in G. Note $9^2 \equiv 81 \equiv 19 \equiv -12 \pmod {31}$. So $9^3 \equiv 9(-12)=108 \equiv -15 \pmod {31}$. Then 9 does not have order 3. Also $9^5 = 9^2 * 9^3 \equiv (-12)(-15)=180 \equiv -6 \pmod {31}$. So 9 does not have order 5. Also $9^10 = 9^5 * 9^5 \equiv (-6)^2 =36 \equiv 5 \pmod {31}$. Then 9 does not have order 10. Therefore it cannot be in the subgroup of of order 10.
To work modulo a small prime like $31$ is not at all tedious. Just start with some positive integer and write down its powers successively, reducing modulo $31$ each time. With luck, you will have chosen a primitive residue, and you get all $30$ nonzero residues. This gives you a log table for multiplication modulo $31$. You see immediately that $2$ is not a good choice, since $2^5\equiv1\pmod{31}$. Try $3$: the powers are $1$, $3$, $9$, $27$, $19$, $26$, $16$, etc. With the full list, you will see that $9$ is not a cube, so that the polynomial is indeed irreducible (cubic polynomial without a root). With your list in front of you, you can do all sorts of multiplicative computations with hardly a thought, for instance finding a high power of any residue at all.