# Squares in Polynomial Quotient Ring $(\mathbb{Z}/2)[x] /(x^{17} - 1)$

Let $R = (\mathbb{Z}/2)[x]$ and $I$ the ideal generated by the polynomial $x^{17} - 1$. Is there a non-zero element of $R/I$ s.t. the square of that element is zero in $R/I$? I know a quotient ring $R/P$ is an integral domain iff $P$ is a prime ideal, and since $R$ is a PID, the prime elements are the irreducible, but $x^{17} - 1$ is reducible in $R$, hence $R/I$ is not an integral domain. So there are nonzero $a,b \in R/I$ s.t. $ab = 0$, but $b$ might not be $a$, and that's where I got stuck. I've also tried finding $f \in R$ s.t. $f^2 = g \cdot (x^{17} - 1)$, but don't know how to attack this methodically.

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Hint: $\:h\ |\ f^2\ \Rightarrow\ h\ |\ f\:$ since $h = x^{17}-1$ is squarefree in $\mathbb Z/2[x],\:$ by $\:(h,h') = (x^{17}-1,x^{16}) = 1\:.\$ See also this post for many characterizations of squarefree elements.