, 2021

Solution:

Let $r = \sqrt{\sqrt{a} + \sqrt{b}}$ and $s = \sqrt{\sqrt{a} - \sqrt{b}}$. We know $r^{2} + s^{2} = 2 \sqrt{a}$ and $r^{2} - s^{2} = 2 \sqrt{b}$. If $r + s$ is an integer $k$, then
$$\sqrt{a} = \frac{r^{2} + s^{2}}{2} = \frac{(r + s)^{2} + (r - s)^{2}}{4} = \frac{k^{2} + 4b / k^{2}}{4},$$
which is rational. Recall that since $a$ is a positive integer, $\sqrt{a}$ is either an integer or an irrational number. (Proof: if $\sqrt{a} = p / q$ for relatively prime positive integers $p, q$, then $p^{2} / q^{2}$ is an integer, which implies $q = 1$.) Thus the answer to part (a) is yes.

The answer to part (b) is no because
$$(2 \pm \sqrt{2})^{2} = 6 \pm \sqrt{32}$$
meaning that setting $a = 36$ and $b = 32$ is a counterexample.

Solution:

Second solution for part (a): Squaring $\sqrt{\sqrt{a} + \sqrt{b}} + \sqrt{\sqrt{a} - \sqrt{b}}$, we see that $2 \sqrt{a} + 2 \sqrt{a - b}$ is an integer. Hence $\sqrt{a - b} = m - \sqrt{a}$ for some rational number $m$. Squaring both sides of this, we see that $a - b = m^{2} - 2m \sqrt{a} + a$, so $\sqrt{a} = \frac{m^{2} + b}{2m}$, a rational number. As in the first solution, it follows that $\sqrt{a}$ is an integer.