Square root of $-1$ in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$ Does there exist a square root of $-1$ in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$?
Now, any element in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$ is of the form $(a+ib)+e(c+id)$ where $i^2=-1$ and $e^2=0$. Now, setting $a=c=d=0$ and $b=1$, I think we would get a square of $-1$. Is this right? Any hints? Thanks beforehand.
 A: That a root exists is immediate from Hensel's Lemma (Newton's method). It is instructive to go futher and compute it by lifting the obvious root $\,x^2\equiv -1\pmod{p\! =\! x^2\!+\!1}\,$ up to $\!\bmod {p^2}.\,$ First we use the general Hensel lifting formula, then we solve it without assuming any knowledge of Hensel.
Note $\  y_1 \equiv\, x\,$ is a root of $\,f(y) = y^2+1\equiv 0\pmod{\!x^2\!+\!1}$
$\!\!\begin{align}{\rm thus}\ \ y_2 &\equiv\, y_1 - f(y_1)\left[\dfrac{1}{f'(y_1)}\bmod x^2\!+\!1\right]\\[.2em]
  &\equiv\, x - (x^2\!+\!1)\left[\,\dfrac{\color{#c00}1}{2\color{#c00}x}\ \ \, \bmod x^2\!+\!1 \right]\\[.2em]
&\equiv\, x - (x^2\!+\!1)\left[\dfrac{\color{#0a0}{-x}}{2}\right]\,\ \ {\rm by}\,\ \ \color{#c00}{\dfrac{1}x}\equiv\dfrac{-x^2}x\equiv \color{#0a0}{-x}\!\!\!\pmod{\!x^2\!+\!1}
\end{align}$
hence $\,y\equiv (x^3+3x)/2\,$ is a root of $\,y^2\equiv -1\, \pmod{(x^2\!+\!1)^2},\, $ by Hensel. 

Remark $ $ If Hensel's Lemma is unknown we can instead do the same directly. Let $\,p = x^2\!+\!1$.
$\!\!\bmod p\!:\ y \equiv a+bx\,$ so $\,-1 \equiv y^2\equiv a^2\!-\!b^2+2ab\,x$ $ \iff a=0,\, b=\pm1\iff y\equiv \pm x$
So, wlog we have: $\, y = x \!+\! g p\,$ for some $\,g\in \Bbb R[x].\,$ Lifting it up to $\!\bmod p^2\,$ we get 
$\!\!\bmod p^2\!:\ {-}1\equiv y^2\equiv (x\!+\!gp)^2\equiv x^2\!+\!2xpg\!$ $\iff\! 2xpg\equiv -p\!$ $\iff\! \bmod p\!:\ g\equiv\large\frac{\color{#c00}{-1}}{2\color{#c00}x}\equiv \frac{\color{#0a0}x}2 $
