27. Find all natural numbers $a$ and $b$ such that $\sqrt{a}-\sqrt{b}$ is a root of the equation $x^{2}+a x-b=0$.
- Answer: $a=2, b=1$. Substitute into our equation the expression $\sqrt{a}-\sqrt{b}$. After some simple algebraic transformations, we get the equality:

$$
\sqrt{b}=\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+2}
$$

Now, square this equality and we get the expression $(4 b-2 a) \sqrt{a}=$ $c$, where $c$ is an integer. It follows that either $a=2 b$, or $\sqrt{a}-$ is a rational number. We will consider the first case later, but from the second observation, it immediately follows that $\sqrt{a}$ is an integer. In this case, $\sqrt{b}$ is also rational, and therefore - an integer. Then, denoting by $a_{1}$ the value of $\sqrt{a}$ (an integer), we get that $a_{1}\left(a_{1}+1\right)$ must be divisible by $a_{1}+2$. But this is impossible, since $a_{1}+1$ is coprime with $a_{1}+2$, and $a_{1}$ is simply smaller. Therefore, it must be that $a=2 b$. Substituting this into the original equation and solving it for $\sqrt{b}$, we find that $\sqrt{b}=1$. That is, we get that $b=1$ and $a=2$. This set works.

(D. Maksimov)