= Solution
The set $\sqrt I$ contains zero. If $x^m\in I$, then $(-x)^m=(-1)^mx^m\in I$, giving closure under additive inverses; also $(rx)^m=r^mx^m\in I$ for every $r\in R$.
To prove closure under addition, let $x^m,y^n\in I$. Expand $(x+y)^{m+n-1}$ by the <binomial theorem>. In every monomial $x^jy^{m+n-1-j}$, either $j\geq m$ or $m+n-1-j\geq n$. Thus each term contains a factor $x^m$ or $y^n$ and belongs to $I$. Hence $(x+y)^{m+n-1}\in I$, so $x+y\in\sqrt I$. The additive subgroup and absorption properties are proved, giving
$$
\boxed{\sqrt I\text{ is an ideal of }R.}
$$
This <ideal> is the <radical of an ideal>. The finite exponent for a sum uses commutativity of the ring.
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