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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.


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.


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.

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$\dfrac{p^2}{q^2} = 2a+ \sqrt{a^2-b}$ Since, $GCD(p,q)=1$ , you can get $a^2-b$ has a square. For the second one, you have to do some manipulation within! – Inceptio Mar 3 '13 at 5:13
You left out a factor of $2$ in the last display. – Gerry Myerson Mar 3 '13 at 5:18
Well, that was a typo! It has to be $2a+ 2\sqrt{a^2-b}$ – Inceptio Mar 3 '13 at 5:20
@GerryMyerson : Now, it seems pretty obvious. Dividing both sides by 4, gives a square of a rational number($\dfrac{p^2}{(2q)^2}$). Which gives the second expression. – Inceptio Mar 3 '13 at 5:23
For some reason I kept missing that factor of 2 every time I attempted the problem. Now that it has been pointed out, the rest falls into place easily. – GovEcon Mar 3 '13 at 5:29
up vote 3 down vote accepted

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."

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What if a and b are themselves rational as well? – Sawarnik Feb 4 '14 at 12:53

Try to square $\sqrt{a+\sqrt b}+\sqrt{a-\sqrt b}$.

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That's precisely what OP did to get from the next-to-last line to the last one. It's just that the attempt wasn't wholly successful. – Gerry Myerson Mar 3 '13 at 5:19

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