Proving $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}\iff a^{2}-b$ is a square 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 ?

Let $F$ be a field of charcteristic different from 2. Let $a$ and $b$ be elements of the field $F$ with $b$ not a square in F. Prove that a necessary and sufficient condition for $\sqrt{a+\sqrt{b}}={\sqrt{m}+\sqrt{n}}$ for $m,n\in F$ is that $a^2-b$ is a square in $F.$
Solution.
$\Rightarrow:$ Suppose that $a^2-b$ is a square in $F$. Then $\sqrt{a^2-b}\in F.$ Let 
$$m= \frac{a+\sqrt{a^2-b}}{2}$$ and $$n= \frac{a-\sqrt{a^2-b}}{2}.$$ Then $n,m\in F$ because $\textrm{char}\, F\neq 0.$ 
$\Leftarrow:$ Now 
$$m =\frac{a+\sqrt{a^2-b}}{2} = \frac{(a+\sqrt{b})+2\sqrt{a^2-b}+(a-\sqrt{b})}{4} = \left(   \frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2} \right)^2,$$ this means that $$\sqrt{m}=\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}.$$
Also
$$n =\frac{a-\sqrt{a^2-b}}{2} = \frac{(a+\sqrt{b})-2\sqrt{a^2-b}+(a-\sqrt{b})}{4} = \left(   \frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2} \right)^2,$$ this means that $$\sqrt{n}=\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2}.$$
Thus $$\sqrt{m}+\sqrt{n} =\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}+\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2}  = \sqrt{a+\sqrt{b}}.$$
