Show $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$ (inverting unit + nilpotent) Show that $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$.
Once you have found the inverse like here, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = $1$ imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?
 A: Write
$$
4x^2+6x+3 = 3(4x^2+2x+1) = 3((2x)^2+(2x)+1) = 3 \frac{(2x)^3-1}{2x-1} = \frac{-3}{2x-1}
$$
Therefore,
$$
\frac{1}{4x^2+6x+3} = \frac{2x-1}{-3} = 3(1-2x) = 3-6x = 2x+3
$$
A: If $R$ is a commutative ring: the units in $R[x]$ are the polynomials whose constant term is a unit, and whose higher order coefficients are nilpotent. You can apply this directly to your example.
A: Hint: As in the hinted paper, a possible ansatz would be
$(4x^2+6x+3) (ax+b) = 4ax^3+(4b+6a)x^2+ (6b+3a)x+3b=1$.
This requires $4a\equiv 0\mod 8$ (so $a$ must be even), $4b+6a\equiv 0\mod 8$, and $6b+3a\equiv 0\mod 8$ and $3b\equiv 1\mod 8$ (so $b=3$).
The cases left are $a$ even with $b=3$.
A: To find an inverse polynomial for that holds $p(x)(4x^2+6x+3)=1$ so it has to be $3y=1\mod 8$ [edit: For more context on $y$, see the comments below]. So $y=3$ and the polynomial might look like this:
$p(x)=(ax+3)$
Then $(4x^2+6x+3)(ax+3)=4ax^3+(6a+12)x^2+(3a+18)x+9$. Now it has to be $4a\equiv 0\mod 8$ and $6a+12\equiv 0\mod 8$ and $3a+18\equiv 0\mod 8$.
Is there such an $a$?.
Yes indeed. For $a=2$ we have $8\equiv 0\mod 8$
$24\equiv 0\mod 8$ and $24\equiv 0\mod 8$.
If we would fail to find this $a$ in this step, we would have to try with $p(x)=(ax^2+bx+3)$ and proceed as above, which gets more and more complicated.
So it is $(4x^2+6x+3)(2x+3)\equiv 1\mod 8$
A: By simpler multiples,  to invert $\, a - f\,$ where $\,a\,$ is invertible, say $\,\color{#0a0}{ab = 1},$ and $\,f\,$ is nilpotent $\color{#c00}{f^n = 0},\,$ we  invert its simpler multiple $\, a^n-\color{#c00}{f^n} = \color{#0a0}{a^n},\,$ with obvious  inverse $\,\color{#0a0}{b^n},\,$ explicitly
$\ \ \ \ \ \ \ \ \ \color{#0a0}{ab=1},\, \color{#c00}{f^{\large n} = 0}\ \Rightarrow\ \overbrace{\dfrac{1}{a-f} = \dfrac{a^{\large n-1}\!+\!\cdots\! +\! f^{\large n-1}}{\!\!\!\!\!\color{#0a0}{a^{\large n}}-\color{#c00}{f^{\large n}}}}^{\large \text{check via cross multiply}} =\, \color{#0a0}{b^{\large n}}(a^{\large n-1}+\cdots + f^{\large n-1})$
$\!\begin{align}{\rm so}\ \ &\color{#0a0}{3(3)=1},\, \color{#c00}{f^{\large 3} = 0}\ \Rightarrow\ \dfrac{1}{3-f} = \dfrac{3^{\large 2}+\,3f\,+\, f^{\large 2^{\phantom{|^{|^.}}\!\!\!\!\!}}}{\color{#0a0}{3^{\large 3}-\color{#c00}{f^{\large 3}}}}\  \ \, =\, \ \ \color{#0a0}{3^{\large 3}}(3^{\large 2}\! +3f + f^{\large 2}) = \bbox[5px,border:1px solid #c00]{2x+3}\\[.1em] 
&{\rm because}\ \ \ \color{#c00}{2^{\large 3}\mid f^{\large 3}}\ \ {\rm by}\ \ 2\mid f = -6x-4x^2,\,\ {\rm to\ invert}\ \ 3\!-\!f = 3\!+\!6x\!+\!4x^2 \in  \smash[b]{\Bbb Z_{\large \color{#c00}{2^{\Large 3}}}}\end{align}$
Generally it is easy to prove that a polynomial is a unit iff its constant term is a unit and all other coefficients are nilpotent (the method of proof there can be made constructive - similar to above).
This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses. Analogous methods may be employed for for computing remainders via modular arithmetic.
