Let $R$ be a principal ideal domain and $x$ a non-zero element of $R$. We define a new $\operatorname{ring} R^{\prime}$ as follows. We define an equivalence relation $\sim$ on $R \times\left\{x^{n} \mid n \in \mathbb{Z}_{\geqslant 0}\right\}$ by

$$\left(r, x^{n}\right) \sim\left(r^{\prime}, x^{n^{\prime}}\right)$$

if and only if $x^{n^{\prime}} r=x^{n} r^{\prime}$. The underlying set of $R^{\prime}$ is the set of $\sim$-equivalence classes. We define addition on $R^{\prime}$ by

$$\left[\left(r, x^{n}\right)\right]+\left[\left(r^{\prime}, x^{n^{\prime}}\right)\right]=\left[\left(x^{n^{\prime}} r+x^{n} r^{\prime}, x^{n+n^{\prime}}\right)\right]$$

and multiplication by $\left[\left(r, x^{n}\right)\right]\left[\left(r^{\prime}, x^{n^{\prime}}\right)\right]=\left[\left(r r^{\prime}, x^{n+n^{\prime}}\right)\right]$.

(a) Show that $R^{\prime}$ is a well defined ring.

(b) Prove that $R^{\prime}$ is a principal ideal domain.