Function

Set this as r.

$r = \sqrt{x - \sqrt{x}}$

${r}^{2} = x - \sqrt{x}$.

We see that because x is an integer, and r is an interger, then x must be a perfect square.

Suppose $x={a}^{2}, a\ge0$

${r}^{2}={a}^{2}-a$.

Correct me if I'm wrong, but I don't think this is possible. 

We see a is larger than r, because you subtract from a larger number to get a smaller number.

Lets suppose a general case where a = r + n, where n > 0, because we know a > r.

Lets compare these two. 

When comparing, and dealing with positives like we are now we generally use if $\frac{a}{b}>1$, then $a>b$, if the opposite the opposite is true.

So we want to see if $\frac{{r}^{2}}{{(r+n)}^{2}-(r+n)}$ is greater than 1 or not.

Simplifying, $\frac{{r}^{2}}{{r}^{2}+rn}$. We see that this is always less than one, because r and n and both positive, so therefore ${r}^{2}<{(r+n)}^{2}-(r+n)$, so therefore substituting a back in, we see ${r}^{2}<{a}^2-a$, so it can't be equal.

So unless you are counting 0, as a 3 digit number, there is no solution.