\documentclass[12pt]{article}
\usepackage{amsmath,amssymb,amsfonts}
\begin{document}
Let $F$ be a free group and $x,y$ be two distinct elements of a
free generating set, then $[x,y]^n$ is not a product of two
squares in $F$, and it is the product of three squares. We give a
short combinatorial proof.
\end{document}