TITLE: Show that $\sqrt{1+t}$ lies in $\mathbb{Z}[1/2][\![t]\!]$
QUESTION [10 upvotes]: Here is an interesting (I thought) question:

Show that $\sqrt{1+t}$ lies in $\mathbb{Z}[1/2][\![t]\!]$

That is, let $f(t)=\sum_{k \ge 0} a_k t^k \in \mathbb{Q}[\![t]\!]$ be the unique power series with $f(t)^2=1+t$ and $f(0)=1$, and show that each $a_k$ is of the form $b/2^m$ for $b,m \in \mathbb{Z}$. 
Naively I just want to "compare coefficients". That is
$$\begin{align}
f(t)^2 &= (a_0+a_1t+a_2 t^2 + \cdots)(a_0+a_1t+a_2 t^2 + \cdots) \\
     &= a_0^2 + (2a_0 a_1)t + \cdots \\
     &= 1 + t
\end{align}$$
Thus given knowledge of $a_k$ it should, in principle, be possible to determine $a_{k+1}$ via $\sum_{i+j=k+1}a_i a_j=0$, but this didn't seem to work out that well. 
In fact the question tells you (one method) to do it - show that $a_k = b_{k-1} + b_k$ where $b_k = \binom{2k}{k}/(-4)^k$. I'm not quite sure how I should do that. Induction seems to be the obvious approach. Recalling that $\sum_{i+j=k+1}a_i a_j=0$ one then inductively knows all terms apart from $a_{k+1}$, but I couldn't figure out how to manipulate the $b_k$'s appropriately. 
Can anyone offer any hints?
Furthermore - can one extend this method to show that $(1+t)^{1/n}$ lies in $\mathbb{Z}[1/n][\![t]\!]$?

REPLY [8 votes]: I had to solve (something similar to) this for a homework assignment this past semester, so I'll post a little about how I solved it.
Suppose we have some $f=a_0+a_1t+\cdots$ such that $f^2=1+t$. Then we have the infinitely many equations
\begin{align}
a_0^2 &= 1\\
2a_0a_1&=1\\
a_0a_2+a_1a_1+a_2a_0&=0\\
a_0a_3+a_1a_2+a_2a_1+a_3a_0&=0\\
\vdots\\
\sum_{i=0}^{j}a_ia_{j-i} &= 0\\
\vdots
\end{align}
Obvious choices for $a_0$ and $a_1$ are 1 and 1/2. As you mentioned in your original post, knowing $a_i$ for $i=0,1\dots,n$ gives you knowledge of $a_{n+1}$, so (starting over) defining $a_0=1,\,a_1=1/2,$ and
$$
a_n=-\frac{1}{2}\sum_{k=1}^{n-1}a_ka_{n-k}
$$
(which is just the $(n+1)$th equation above solved for $a_n$) sets you up for an easy inductive proof that $f^2=t+1$.