Show $100+t$ is reducible in $\mathbb{Z}[[t]]$ $\mathbb{Z}[[t]]$ is the ring of formal power series with integer coefficients, and the problem asks to show that $100+t$ is reducible. This means I need to find two power series, $\sum_{i=0}^\infty a_i t^i$ and $\sum_{j=0}^\infty b_j t^j$ with $a_i,b_j\in\mathbb{Z}$, that when multiplied give $100+t$. Since the integers are an integral domain, I know that neither of these power series may be finite, but I'm having a really hard time figuring out how to come up with two series that will telescope properly. Any suggestions?
 A: You want to find $(a_i)$ and $(b_i)$ such that $(\sum_i a_i t^i) (\sum_i b_i t^i) = 100+t$. Multiply this out and compare the coefficients:
$$a_0 b_0 = 100\\
a_0 b_1+a_1 b_0 = 1\\
a_0 b_n + a_1 b_{n-1} + \ldots + a_{n-1} b_1 + a_n b_0 = 0 \quad \text{for } n \geq 2
$$
The second equation tells you that you should choose $a_0$ and $b_0$ coprime, so let $a_0 := 2^2$ and $b_0 := 5^2$. From this you can inductively construct all the coefficients $(a_i)$ and $(b_i)$, but I let you do the details.
A: First, solve it modulo $t$. Then modulo $t^2$. Then....
Splitting $100 = 25 \cdot 4$ is the only choice, because that is the only way to factor $100$ into relatively prime parts. this is important for reasons that will be clear if you try to apply the method below)
So, I've chosen


*

*$f \equiv 25 \pmod t$

*$g \equiv 4  \pmod t$


To set some terminology, let


*

*$f_n$ be the truncation of $f$ to degree $n$

*$g_n$ be the truncation of $g$ to degree $n$

*$a_n$ be the coefficient of $t^n$ in $f$

*$b_n$ be the coefficient of $t^n$ in $g$


Now, to get the coefficient on $t$:


*

*$(f_0 + t a_1) (g_0 + t b_1) = 100 + t (4 a_1 + 25 b_1) \pmod {t^2}$


So I need do an extended GCD and decide upon $a_1 = -6$ and $b_1 = 1$.
Now, induct. By hypothesis, we may assume
$$ f_{n-1} g_{n-1} \equiv 100 + t \pmod {t^n} $$
and going up a degree gives
$$f_{n-1} g_{n-1} \equiv 100 + t + t^n c_n \pmod{ t^{n+1}} $$
for some value of $c_n$.
$$ \begin{align}
f_n g_n &= (f_{n-1} + a_n t^n) (g_{n-1} + b_n t^n) & \pmod {t^{n+1}}
\\ &= 100 + t + t^n (c_n + a_n g_{n-1}(0) + b_n f_{n-1}(0)) & \pmod{ t^{n+1}}
\\ &= 100 + t + t^n (c_n + 4 a_n  + 25 b_n ) & \pmod{ t^{n+1}}
\end{align} $$
From which it's clear that we can solve for $a_n$ and $b_n$. e.g. $a_n = 6 c_n$ and $b_n = -c_n$
Since we obtain $f_n$ and $g_n$ for all $n$, we get a solution for $f$ and $g$.
EDIT: Notice at this point, we can observe that one of the solutions has $a_n = -6 b_n$ for $n > 0$, and so we could write $100 = (25 + 6 t h(x)) (4 - t h(x))$, and then segue into Bill Dubuque's method.
A: To split $\rm\,ab + t\,$ with $\rm\,\gcd(a,b)=1,\,$ write $\rm\ \color{#c00}{\bf 1} = ad\!-\!bc\,$ by Bezout's GCD identity.
Recall the Catalan series $\rm\, C(x)\in \mathbb Z[[x]]\,$ satisfies $\rm\,\color{#090}{C(x) - x\:C(x)^2\! = 1},\,$ so with $\rm\,e = cd$
$\ \ \  \rm (a - ct\:C(et))\:(b+dt\:C(et))\ =\ ab + \color{#c00}{\bf 1}\cdot t\:(\color{#090}{C(et) - et\:C(et)^2})\: =\ ab + t$
Remark $\ $ Below is the method that I used to discover this empirically.
By undetermined coefficients and an OEIS lookup one deduces empirically that
$$\rm\ 100 + t\: =\: (4 + f(t))\:(25 - 6\: f(t))\quad where$$
$$\rm f(t)\: =\: t\:C(6t)\: =\: t\sum_{n\:=\:1}^{\infty}\:  C_n (6t)^n\ =\ t + 6\: t^2 + 72\: t^3 + 1080\: t^4 + 18144\: t^5 +\:\ldots $$
where $\rm\displaystyle\ \ C_n =\: {\rm Catalan}(n)\:=\:\dfrac{1}{n+1}{2n\choose n}\ =\ \frac{(2n)!}{n!\:(n+1)!}\ $ with generating function 
$$\smash[b]{\rm C(t)\: =\ \sum_{n\:=\:0}^{\infty} C_n t^n\ =\ {\dfrac{1-\sqrt{1-4t}}{2t}}}$$
This yields the closed form
$$\begin{eqnarray}{}\rm 100 + t\, &=&\rm\ \  (4\ +\ t\:C(6t))\ \ \ \ (25\ -\ 6t\:C(6t)) \\[0.4em]
&=&\rm \dfrac{49 - \sqrt{1-24\:t}}{12}\ \dfrac{49 + \sqrt{1-24\:t}}{2}\\
\end{eqnarray}$$
In fact, employing the Catalan functional equation $\rm\ \color{#090}{C(x) - x\:C(x)^2 = 1}\:$  we confirm
$$\rm\ (4 + t\:C(6t))\:(25-6t\:C(6t))\ =\  100 + t\:(\color{#090}{C(6t)-6t\:C(6t)^2})\ =\ 100 + t\quad QED $$
