Multiplicative inverse of $4x+3$ in the ring $ \mathbb Z_{11}[x]/(x^2+1)$

Question

How to find the multiplicative inverse of $4x+3$ in the ring $ \mathbb Z_{11}[x]/(x^2+1)$?

I understand that any element belonging to the ring $ \mathbb Z_{11}[x]/(x^2+1)$ is of the form $ax+b$ where $a,b \in \mathbb Z_{11}$. How to proceed?

Answer 1

Note that $x^2=-1$, thus $$ \frac1{4x+3}=\frac{4x-3}{(4x+3)(4x-3)}=\frac{3-4x}3=1+6x $$ (I used that $4=3^{-1}\bmod 11$).

Answer 2

Use the Euclidean algorithm in $\mathbf Z/11\mathbf Z$ to obtain a Bézout's relation between $x^2+1$ and $4x+3$:

\begin{alignat}{6} &3x+6 \\ &———&~\\ 4x+3\;\Bigl(\;&\phantom{-}x^2\phantom{{}+2x}{}+1\\ &{-x^2}+2x&\\ &~~~~~————&~\\ &~~~\phantom{——}2x+1 \\ &~~~\phantom{—}-2x+1 \\ &~~~~~~\phantom{—}———&~\\ &~~~~~~~\phantom{———}5& \end{alignat}

Thus we have \begin{align}x^2+1=(3x+6)(4x+3)+5&\iff(3x+6)(4x+3)\equiv-5\pmod {x^2+1}\\ &\iff 2(3x+6)(4x+3)\equiv 1\pmod{x^2+1}, \end{align} and the inverse is $$(4x+3)^{-1}=2(3x+6)=6x+1.$$