Let $R$ be a commutative ring with identity. Is there any characterization for invertible elements of $R[x,x^{1}]$ ?
I have read about this somewhere. I think it was as follows: $\sum_{i=n}^n a_ix^i \in R[x,x^{1}]$ is invertible iff $\sum a_i^2$ is invertible in $R$ and for all $i \not = j$, $a_ia_j$ is nilpotent. 


Thinking geometrically in terms of the map ${\rm{Spec}}(R[x,1/x]) \rightarrow {\rm{Spec}}(R)$ and noting that being a unit amounts to being nonzero in the residue field at every prime, an element $f = \sum a_i x^i \in R[x,1/x]$ is a unit if and only if it has unit restriction to every fiber, which is to say that for every prime ideal $P$ of $R$ (with residue field $k(P)$) the image $f(P) := \sum a_i(P) x^i$ in $k(P)[x,1/x]$ is a unit. But since $k(P)$ is a field, this latter condition is exactly that $f(P)$ is a $k(P)^{\times}$multiple of a power of $x$. That is, there is exactly one $i$ (depending perhaps on $P$) such that $a_i(P) \ne 0$, which can be equivalently expressed as the condition that $a_i(P)a_j(P) = 0$ in $k(P)$ for all $i \ne j$ and $\sum a_i(P) \ne 0$ in $k(P)$. Varying over all $P$, this necessary and sufficient condition says exactly that (1) $a_i a_j$ is nilpotent in $R$ when $i \ne j$ and (2) $\sum a_i \in R^{\times}$. In the presence of (1), squaring the sum in (2) (which has no effect on whether or not it is a unit) and noting that adding a nilpotent element has no effect on being a unit shows that (2) can be replaced with (2') $\sum a_i^2 \in R^{\times}$ (thereby recovering the formulation in shatich's answer). 


Invertible elements of Laurent algebras, and more generally of algebras of torsionfree, cancellable commutative monoids, are characterised in Theorem 11.3 and Corollary 11.4 of Gilmer's Commutative Semigroup Rings (Chicago Lectures in Mathematics, 1984). The proofs given there are quite accessible. 

