Can a finite sum of square roots be an integer?

I think this link is a pretty good answer to your question. However, it might be at a level which is too advanced for you, since this is a pretty natural question to ask relatively early on in life, but it takes some significantly more difficult mathematics to prove.

The direct, yes/no answer to the question is "Yes, but only if the numbers inside the square roots were already perfect squares," or equivalently "If you've already done all the simplifying that you can do, then no."

At least there's an elementary way to see that if $\sqrt{a} + \sqrt{b}$ is an integer, then $a$ and $b$ are perfect squares.

Suppose $\sqrt{a} + \sqrt{b} = c\in\mathbb{Z}.$ If $c=0$ the result is trivial. Otherwise, squaring both sides we get that $$a + b + 2\sqrt{ab} = c^2$$ and therefore $ab$ must be a perfect square. Let's say $ab = d^2$. Then $a=\frac{d^2}{b}$ and \begin{align*}\frac{d}{\sqrt{b}} + \sqrt{b} &= c\\ d + b &= c\sqrt{b}, \end{align*} so $b$ is a perfect square, and $a$ must be as well.

Suppose that $a,b,\sqrt a+\sqrt b\in\mathbb Z$.

$(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b\in\mathbb Z$. Since $\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}\in\mathbb Q$. Therefore, $\sqrt a-\sqrt b$ is an algebraic integer and rational; thus, $\sqrt a-\sqrt b\in\mathbb Z$.

Next, $(\sqrt a+\sqrt b)+(\sqrt a-\sqrt b)=2\sqrt a\in\mathbb Z$ and $(\sqrt a+\sqrt b)-(\sqrt a-\sqrt b)=2\sqrt b\in\mathbb Z$. Thus, $\sqrt a$ and $\sqrt b$ are algebraic integers and rational, therefore $\sqrt a,\sqrt b\in\mathbb Z$.

Thus, $a,b,\sqrt a+\sqrt b\in\mathbb Z\Rightarrow\sqrt a,\sqrt b\in\mathbb Z$