Conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational Motivation
I am working on one of the questions from Hardy's Course of Pure Mathematics and was wondering if I could get some assistance on where to go next in my proof. I have attempted rearranging the expression in numerous ways from the step I am at, but seem to get no-where.
Question
If $a^2-b>0$, then the necessary and sufficient conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational are $a^2-b$ and $\dfrac{1}{2} (a+ \sqrt{a^2-b})$ be squares of rational numbers.
Attempt
Suppose that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational. Then it can be written as the ratio of two integers, p and q, that have no common factor. Write this as:
$\dfrac{p}{q}=\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$
Then by squaring both sides we have:
$\dfrac{p^2}{q^2} = (\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}})^2=2a+ 2\sqrt{a^2-b}$
-Note sure where to go from here.

The assertion that "If $a^2−b>0$, then the necessary and sufficient conditions that $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is rational are $a^2−b$ and $\frac{1}{2}(a+\sqrt{a^2-b})$ be squares of rational numbers." is false. Take $a=2-\sqrt[4]{2}, b=4-4\sqrt[4]{2}, a^2-b=\sqrt{2}, \sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}=2$.
I believe the assertion should read: "If $a^2−b>0$, then the necessary and sufficient condition that $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is rational is that $\frac{1}{2}(a+\sqrt{a^2-b})$ is a square of a rational number."
