Finding the square root of $6-4\sqrt{2}$ I found this standupmaths video on YouTube about the A4 paper puzzle.
I really liked the puzzle and managed to get the answer by using a calculator. However, the answer (which I won't spoil), led me to think that the equation to solve it might simplify - which it does.
In the middle of the simplification, I got this expression:
$$\sqrt{6-4\sqrt{2}}$$
which for other reasons I suspected to be equal to:
$$\ 2-\sqrt{2}$$
I tried squaring the above and, sure enough, it does give:
$$6-4\sqrt{2}$$
My question is, how would I have been able to find the square root of
$$6-4\sqrt{2}$$
if I hadn't been able to guess at it? Is there a standard technique? I've tried looking on the web but don't really even know what to search for.
 A: One way of doing this would be to guess that 
$$\sqrt{6-4\sqrt{2}} = a+b\sqrt2$$
for some $a,b \in \mathbb Z$. Squaring both sides, we get
$$6-4\sqrt2 = a^2+2b^2 + 2ab\sqrt2$$
which implies that $6=a^2+2b^2$ and $-4 = 2ab$. Since $a$ and $b$ are integers, there are two solutions: $(a,b) = (2,-1)$, and $(a,b) = (-2,1)$, and we reject the latter.
A: The numbers of the form $a+b\sqrt{2}$ share many arithmetic properties with the integers. That's why you might suspect an answer of that form to the question. So try one:
$$
(a+b\sqrt{2})^2 = a^2 + 2b^2 + 2ab\sqrt{2} = 6-4\sqrt{2}.
$$
Then it's easy to finish.
You can find out more on wikipedia (https://en.wikipedia.org/wiki/Quadratic_integer) but you'd have to know to search for "quadratic integer" to find that page.
A: We have to assume that the nested radical can be rewritten as the sum of two other radicals (surds).
$\sqrt{6-4\sqrt{2}}=\sqrt{d}+\sqrt{e}$
Squaring both sides gives us $$6-4\sqrt{2}=d+e+2\sqrt{de}$$
This can be solved by finding $2$ numbers that sum to $6$ and multiply to $4^2\cdot 2/4=8$
($2\sqrt{de}=4\sqrt{2}\rightarrow de=8$)
Numbers $4$ and $2$ work, so $$\sqrt{6-4\sqrt{2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}$$
A: set $$(a+b\sqrt{2})^2=6-4\sqrt{2}$$ and find the rational numbers $a$ and $b$
