Let $I$ be a radical ideal of a regular local ring $S$. Put $R:=S/I$. Let $I^{(n)}$ be the $n$-th symbolic power of $I$. It is well-known that $I^n \subseteq I^{(n)}$.
Is it true that $I/I^2$ is $R$-free if and only if $I/I^{(2)}$ is $R$-free?
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In your case, when $I$ is reduced, one can show that $I/I^2$ is free implies $I$ is a complete intersection and then $I^2=I^{(2)}$.
For this, let height of $I$ be $r$. Let $P$ be a minimal prime of height $r$ of $I$. Since $I$ is reduced, we get $I_P=P_P$ and since $R_P$ is a regular local ring of dimension $r$, we get that $I/I^2$ must be a free module of rank $r$. So, $I$ is generated by $r$ elements and thus it is a complete intersection.
I do not know whether the converse is true.
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$\begingroup$ @uno For the converse statement of what Mohan wrote, the Main Theorem of the paper titled A Variation on a Theme of Vasconcelos by Daniel A. Smith (Journal of Algebra, 225, 381–390 (2000)) may be of interest. $\endgroup$ Commented Feb 6 at 0:21