14:3.26 ** Find all natural numbers that are equal to the square of their number of divisors.



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The numbers that satisfy the problem are only 1 and 9.
Let $n>1$ be a number that satisfies the problem, and $n$ has $m \geqslant 1$ positive divisors, then $n=m^{2}$, and a square number has an odd number of positive divisors, so $m$ is odd. Let $m=2 k+1$ (where $k$ is a positive integer), then $n$ has $k$ divisors less than $m$. They are all odd. That is, its $k$ divisors less than $m$ are $1,3, \cdots, 2 k-1$. Therefore, $2 k-1 \mid(2 k+1)^{2}$. Thus, $2 k-1\left|(2 k+1)^{2}-(2 k-1)^{2}, 2 k-1\right| 4$. So $k=1, n=9$.