There are integers satisfying an integer?

$\frac{x-3\sqrt{x}+2}{\sqrt{x}}$ $=\sqrt{x}-3+\frac{2}{\sqrt{x}}$

For this to be an integer, $x$ must be a perfect square and $\sqrt{x}$ must divide $2$

Factors of $2 : 1,2,-1,-2$

So, $\sqrt{x}=1,2,-1,-2$ or $x=1,4$

Answer is sum of values of $x$ , that is, $\boxed{5}$

I think the answer should be $3$ , with solutions $-2, 1, 4$ . (But would probably be better if you specified "positive integer" instead.

It's an integer iff $(1 + \frac{2}{x}) \sqrt{x}$ is an integer. This occurs when, $\sqrt{x}$ in an integer and so is $\frac{2}{\sqrt{x}}$ , e.g $x = 1, 4$ . Or when $(1 + \frac{2}{x}) = 0$ . In this case $\sqrt{x} = \sqrt{2} i$ and is an imaginary number.

Also note, $\frac{x - 3\sqrt{x} + 2}{\sqrt{x}} = \frac{(\sqrt{x}-1)(\sqrt{x}-2)}{\sqrt{x}} = \frac{(y-1)(y-2)}{y}$ . Which is $0$ when $y = 1, 2$ .