Let $A$ be a $m\times n$ real matrix. Then $B:=A^TA$ is an $n\times n$ symmetric matrix. Is the converse true?
More precisely, given any $n\times n$ symmetric matrix $B$ and any positive integer $m$, can $B$ be written as $A^TA$ for some $m\times n$ matrix $A$?
If you consider the set of $1\times1$ real matrices (i.e. the real numbers), then you can see immediately that if $B=-1$ then $A=i$.
However if you allow the entries of $A$ to be complex, then this follows from the special form of the eigendecomposition for symmetric matrices,
$$B=P^TDP=P^T\sqrt{D}^T\sqrt{D}P = (\sqrt{D}P)^T\sqrt{D}P.$$
To ensure that the entries of $A$ were real would require the extra assumption that $B$ was also positive semidefinite (all eigenvalues nonnegative).
I guess I should also clarify, in case you meant in your question to fix $m$ prior to searching for an appropriate $A$, the answer is in general no, since the rank of $A^TA$ is upper bounded by $\min(m,n)$, and thus, for example, any $B$ of full rank cannot be constructed when $m < n$.