$\mathbf {The \ Problem \ is}:$ Does $-1$ have a square root in the ring $\frac{\mathbb R[x]}{\langle(x^2+1)^2\rangle}$ ? Justify your assertion .
$\mathbf {My \ approach} :$ Actually, the above quotient ring $R$ is obviously not an integral domain, and $\langle(x^2+1)^2\rangle \subset \langle(x^2+1)\rangle$ and we knnow $\frac{\mathbb R[x]}{\langle(x^2+1)\rangle} \cong \mathbb C$, but I can't get that what is meant by square root of $-1$ in $R$ ?
A little help is greatly appreciated.
On
I will prove that in the ring $R_k=\Bbb R[x]/\big\langle (x^2+1)^k\big\rangle$, $-1$ has a square root. In other words I will exhibit a polynomial $p_k(x)\in \Bbb R[x]$ of degree at most $2k-1$ such that $\big(p_k(x)\big)^2+1$ is divisible by $(x^2+1)^k$. Moreover $p_k(x)$ is unique up to sign change.
First we will prove uniqueness (up to sign change). If $\tilde{p}_k(x)$ is another polynomial of degree at most $2k-1$ such that $\big(\tilde{p}_k(x)\big)^2+1$ is divisible by $(x^2+1)^k$, then
$$\big(p_k(x)-\tilde{p}_k(x)\big)\big(p_k(x)+\tilde{p}_k(x)\big)=\Big(\big(p_k(x)\big)^2+1\Big)-\Big(\big(\tilde{p}_k(x)\big)^2+1\Big)$$
is divisible by $(x^2+1)^k$. Since both $p_k(x)$ and $\tilde{p}_k(x)$ are coprime to $x^2+1$, we conclude that $p_k(x)-\tilde{p}_k(x)$ and $p_k(x)+\tilde{p}_k(x)$ cannot both be divisible by $x^2+1$. Hence $(x^2+1)^k$ must divide exactly one of $p_k(x)-\tilde{p}_k(x)$ and $p_k(x)+\tilde{p}_k(x)$. Since $(x^2+1)^k$ has degree $2k$ whilst $p_k(x)-\tilde{p}_k(x)$ and $p_k(x)+\tilde{p}_k(x)$ are of degrees at most $2k-1$, we must have $p_k(x)-\tilde{p}_k(x)=0$ or $p_k(x)+\tilde{p}_k(x)=0$, proving the assertion.
Now for the existence, we first note that if $p_k(x)$ exists, then $$\big(p_k(x)\big)^2+1=(x^2+1)^kq_k(x)$$ for some $q_k(x)\in\Bbb R[x]$ of degree at most $2k-2$. Taking derivative of the equation above yields $$2p_k(x)p'_k(x)=2kx(x^2+1)^{k-1}q_k(x)+(x^2+1)^kq_k'(x),$$ which is clearly divisible by $(x^2+1)^{k-1}$. Since $x^2+1$ is coprime to $p_k(x)$, we see that $(x^2+1)^{k-1}$ must divide $p'_k(x)$. As $\deg p'_k=\deg p_k -1\le (2k-1)-1=2k-2$ and the degree of $(x^2+1)^{k-1}$ is $2(k-1)=2k-2$, we must have $$p'_k(x)=A_k(x^2+1)^{k-1}$$ for some constant $A_k$. We make a guess that $p_k(x)$ has no constant term so $$p_k(x)=\int_0^x p'_k(t)dt=A_k\int_0^x (t^2+1)^{k-1}dt.$$ Since $(x^2+1)^k$ divides $\big(p_k(x)\big)^2+1$, $\big(p_k(i)\big)^2+1=0$. So we may pick $p_k(i)=i$. That is $$i=p_k(i)=A_k\int_0^i (t^2+1)^{k-1}dt=i A_k\int_0^1 (1-s^2)^{k-1}ds,$$ where $s=-it$. The integral $\int_0^1(1-s^2)^{k-1}ds$ is well known, which is equal to $\frac{(2k-2)!!}{(2k-1)!!}$, and can be calculated inductively on $k$ or with the use of the beta function. That is $A_k=\frac{(2k-1)!!}{(2k-2)!!}$, so $$p_k(x)=\frac{(2k-1)!!}{(2k-2)!!}\int_0^x (t^2+1)^{k-1}dt=\frac{(2k-1)!!}{(2k-2)!!}\sum_{r=1}^{k}\frac{1}{2r-1}\binom{k-1}{r-1}x^{2r-1}.$$ We now need to verify that $\big(p_k(x)\big)^2+1$ is indeed divisible by $(x^2+1)^k$. Since $(x^2+1)^{k-1}$ divides $p'_k(x)$, which means $i$ is a root of the derivative $2p_k(x)p'_k(x)$ of $\big(p_k(x)\big)^2+1$ with multiplicity $k-1$. By construction, $i$ is also a root of $\big(p_k(x)\big)^2+1$. This means $i$ is a root of $\big(p_k(x)\big)^2+1$ with multiplicity $k$. Hence the minimal polynomial $x^2+1$ of $i$ must divide $\big(p_k(x)\big)^2+1$ precisely $k$ times, establishing the claim.
Here is a table of $p_k(x)$ and $q_k(x)$ for $k=1,2,3,4,5$. From this table I conjecture that $q_k(x)$ is an irreducible polynomial in $\Bbb Q[x]$.
$$\begin{array}{|c|c|c|} \hline k&p_k(x)&q_k(x) \\ \hline 1&x&1\\ 2&\frac{x^3+3x}{2}&\frac{x^2+4}{4}\\ 3&\frac{3x^5+10x^3+15x}{8}&\frac{9x^4+33x^2+64}{64}\\ 4&\frac{5x^7+21x^5+35x^3+35x}{16}&\frac{25x^6+110x^4+201x^2+256}{256}\\ 5&\frac{35x^9+180x^7+378x^5+420x^3+315x}{128}&\frac{1225x^8+6475x^6+14235x^4+17305x^2+16384}{16384}\\\hline \end{array} $$
Elements of $\mathbb{R}[x]/\langle (x^2+1)^2\rangle $ are of the form $$\{ a+bx+cx^2+dx^3: x^4=-2x^2-1\}$$ therefore you asked to find an element $a+bx+cx^2+dx^3$ of $\mathbb{R}[x]/\langle (x^2+1)^2\rangle $ such that $$(a+bx+cx^2+dx^3)^2=-1$$ Expand the LHS subject to $x^4=-2x^2-1$, $x^5=-2x^3-x$ and $x^6=-2x^4-x^2=3x^2+1$ to see if there exist real calues for $a,b,c,d$.