Is it true that if some power of an ideal $I$ is primary, then $I$ itself is also a primary ideal?
I do not know whether the above statement is true or there is a counterexample. If one wants to prove it, then the natural way is as follows. Suppose the $n$-th power of $I$ is primary for some $n\geq 2$. Let $ab\in I$ and $a\notin I$. Then one should show that $b\in \sqrt{I}$. Since $ab\in I$, we have $a^nb^n={(ab)}^n\in I^n$ (we are in a commutative ring). If $a^n\notin I^n$, then we get $b^n\in \sqrt{I^n}=\sqrt{I}$, since $I^n$ is primary by assumption. So, in this case, we get $b\in \sqrt{I}$, since $\sqrt{I}$ is a prime ideal as well as $\sqrt{I^n}$, and hence we are done. But, if $a^n\in I^n$, what can we say? (Indeed, the assumption $a\notin I$ does not necessarily imply that $a^n\notin I^n$). So, maybe one can find a counterexample, or somehow finish the proof.