 Solution

Certainly $A\subseteq \sqrt{A}$, by taking exponent one. If $x\in \sqrt{A}$, choose $n\ge 1$ with $x^n\in A$. Primality applied to $x\cdot x^{n-1}$ gives $x\in A$ or $x^{n-1}\in A$. In the latter case repeat; induction reduces to $x\in A$. Thus $\sqrt{A}\subseteq A$ as well, provingEvery prime ideal is therefore a radical ideal.