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

Question

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.

Answer 1

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.