Perfect square in field of formal Laurent series Let $k$ be an algebraically closed field of characteristic not equal to $2$, and consider the field $k(\!(t)\!)$ of formal laurent series over $k$. I'm trying to show that an element $\sum_{i=N}^\infty a_it^i \in k(\!(t)\!)$ with $N\in\mathbb Z$, $a_i\in k$ for $i\geq N$, and $a_N \neq 0$, is a square in $k(\!(t)\!)$ if and only if $N\in 2\mathbb Z$.
If I understand how multiplication works, I think that
$$
\Bigl(\sum_{i=N}^\infty a_it^i \Bigr)^2 = \sum_{i = 2N}^\infty \Bigl( \sum_{j + k= i} a_ja_k \Bigr) t^i.
$$
It's clear from this that every perfect square must have as its first nonzero term, $t$ raised to an even integer. What I don't get from this, though, is how to show that this condition is also sufficient for being a perfect square. Any suggestions?
 A: We can show that any power series of the form $1+tf$ has a square root.  One approach is to expand $\sqrt{1+X}$ as a power series in $X$, and set $X=tf$.
The result follows because we can divide out by a constant times an even power of $t$, and constants have square roots because $k$ is algebraically closed.
A: This direction is where you'll use the fact that $k$ is algebraically closed.
If you want to square a formal Laurent series, what you wrote is mostly right, but with a caveat:
$$ \left( \sum_{i=N}^\infty a_i t^i \right)^2 = \sum_{i=2N}^\infty \left(\sum_{\substack{j+k=i \\ j, k \geq N}} a_j a_k \right) t^i $$
This means that, given a formal Laurent series of the form
$$ \sum_{i=2N}^\infty c_i t^i $$
we can go looking for a square root --
$$ \sum_{i=2N}^\infty c_i t^i = \left( \sum_{n=N}^\infty a_n t^n \right)^2 $$
or equivalently, we can go looking for a sequence of terms $(a_n)_{n=N}^\infty$ such that, whenever $i \geq 2N$,
$$ c_i = \sum_{\substack{j+k=i \\ j, k \geq N}} a_j a_k $$
Let's tackle this bad boy inductively. For $i=2N$, the only pairs $j, k \geq N$ where $j+k = 2N$ are $j, k = N$. So necessarily $c_{2N} = \left(a_N \right)^2$. Since $k$ is algebraically closed, we can solve -- 
$$ a_N = \sqrt{c_{2N}} $$
Remember, our goal is to solve for the entire sequence $(a_n)$. If we can do that, we'll have found a square root of the starting Laurent series. Now let's proceed by strong induction -- assuming that we've defined $(a_n)$ for $n = N, N+1, \cdots, N+n-1$. Now our goal is to show that we can define $a_{N+n}$. We have the relation
$$ c_{2N+n} = \sum_{\substack{j+k = 2N+n \\ j, k \geq N}} a_j a_k $$
You can pick off the highest terms:
$$c_{2N+n} = 2 a_{N}a_{N+n} + \sum_{\substack{j+k = 2N+n \\ j, k > N}} a_j a_k $$
which allows you to solve for $a_{N+n}$ in terms of values that, by induction, are well-defined and in $k$.
A: A Laurent series writes $f(X)=X^{-N}g(X)$ with $g\in k[[X]]$.
If $N=-2m$ then finding a square root for $f$ is equivalent to finding a square root for $g$, that is, $h\in k[[X]]$ such that $g=h^2$.
This can be done step by step as follows: write $g(X)=a_0+a_1X+\cdots$ and search for $h(X)=b_0+b_1X+\cdots$ such that $g=h^2$, that is, $a_0=b_0^2$, $a_1=2b_0b_1$, $a_2=b_1^2+2b_0b_2$, and so on. We get $b_0=\sqrt{a_0}$ (here one uses that $k$ is algebraically closed), $b_1=(2\sqrt{a_0})^{-1}a_1$, $b_2=(2\sqrt{a_0})^{-1}[a_2-(2\sqrt{a_0})^{-2}a_1^2]$, and so on.
