TITLE: Is $t^4+7$ reducible over $\mathbb{Z}_{17}$?
QUESTION [8 upvotes]: Is $f=t^4+7$ reducible over $\mathbb{Z}_{17}$?
Attempt: I checked that $f$ has not roots in $\mathbb{Z}_{17}$, so the only possible factorization is with quadratic factors.
Assuming $f=(t^2+at+b)(t^2+ct+d)$, we have $bd=7$, $a+c=0$ and $ac+b+d=0.$ But it is cumbersome to find if there are solutions for these equations.
I know certain tools to prove the irreducibility of a polynomial over rings like $\mathbb{Z}$ or $\mathbb{Q}$ such as Gauss and Eisenstein's criteria. But over finite rings like $\mathbb{Z}_n$, I don't know how to prove the irreducibility of a polynomial with degree higher than $3$. Finding roots allows me to find linear factors, but I cannot use this technique to find quadratic factors. What kind of tool may I use in this case?

REPLY [4 votes]: Adding another approach.
Let $\alpha$ be a root of $f$ in some extension field of $\mathbb{Z}_{17}$. We know that $\alpha^4=-7=10$. Either brute force checking (won't take long!) or toying with quadratic reciprocity and such shows that $10$ is not a quadratic residue modulo $17$. As $\mathbb{Z}_{17}^*$ is cyclic of order $16$, this means that the order of $10$ is exactly $16$ (if it were a proper factor, $10$ would have to be a square).
This allows us to deduce that $\alpha$ is primitive root of unity of order $64$. Therefore $\alpha$ cannot be a zero of a quadratic polynomial over $\mathbb{Z}_{17}$, because the zeros of such polynomials belong to the field $\mathbb{F}_{289}$ of $17^2=289$ elements. The contradiction comes from the fact that the multiplicative group of $\mathbb{F}_{289}$ is cyclic of order $288$ but $288=2^5\cdot3^2$  is not divisible by $64=2^6$.