I was trying to prove the following:
Let $R$ be a commutative ring. Suppose $I$ is an ideal of $R$. Then $I$ is a radical ideal if and only if for $x\in R$ such that $x^2\in I$ then $x\in I$.
One direction is very easy. If you suppose $I$ is radical then the result is immediate.
The other direction is not so easy. We have to prove that for any $n$, $x^n\in I\Rightarrow x\in I$. I tried to prove it by induction, but there is some trouble when $n$ is odd. I think the problem can be reduced to prove it when $n$ is an odd prime.
But I don't know if this is the best approach. Perhaps it's better to suppose some $x$ is in all prime ideals containing $I$ and try to prove $x\in I$ but at first sight it seems more convoluted to do it that way. So, how can I proceed?
Assume your condition $x^2\in I\implies x\in I$ holds.
Suppose $x^3\in I$. Then $x^4\in I$ since $I$ is an ideal, and $x^4=xx^3$. Therefore $x^2\in I$ (as $(x^2)^2\in I$) and so $x\in I$.
See above for the $x^4\in I$ case.
Suppose $x^5\in I$. Then $x^6\in I$ since $I$ is an ideal, and $x^6=xx^5$. Therefore $x^3\in I$ (as $(x^3)^2\in I$) and we have seen that implies that $x\in I$.
And so on...