This is an exercise for the book Abstract Algebra by Dummit and Foote
(pg. 530):
Let $F$ be a field of characteristic $\neq2$ . Let $a,b\in F$ with $b$
not a square in $F$. Prove $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$ for
some $m,n\in F$ iff $a^{2}-b$ is a square in $F$.
I am having problem proving this claim, I tried to assume $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$
and I naturally squared both sides, to try and get $a^{2}$ I squared
both sides again and then reduced $2b$ from both sides and rearranged
to get $$a^{2}-b=(m+n+2\sqrt{mn})^{2}-2\sqrt{b}(a+\sqrt{b})$$ but
I don't see how I can use it.
Can someone please help me prove this claim ?
Best Answer
$$\sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}$$ $$a+\sqrt{b} = m+n+2\sqrt{mn}$$ Thus $a = m+n$ and $b = 4mn$ as $b$ is not a square . Finally, $$a^2 -b = m^2 + 2mn + n^2 -4mn = (m-n)^2$$.