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 ?

$$\sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}$$ $$a+\sqrt{b} = m+n+2\sqrt{mn}$$ Thus $a = m+n$ and $b = 4mn$ as $b$ is not a square . Finally, $$a^2 -b = m^2 + 2mn + n^2 -4mn = (m-n)^2$$.

$$\sqrt{a+\sqrt b}=\sqrt m + \sqrt n \Rightarrow a+\sqrt b = m+n+2\sqrt{mn}$$ Since $\phi(\alpha+\beta\sqrt b)=\alpha-\beta\sqrt b$ for $\alpha,\beta\in F$ defines an automorphism $\phi\colon F[\sqrt b]\to F[\sqrt b]$ that leaves $F$ fixed, we have that $\phi(\sqrt{mn})= \pm\sqrt{mn}$ because $\phi$ maps the polynomial $X^2-mn$ to itself and can at most interchange its roots. Thus we additionally get $a-\sqrt b=\phi(a+\sqrt b)=\phi(m+n+2\sqrt{mn})$, i.e. $$a-\sqrt b=m+n\pm2\sqrt{mn}.$$ Since $\sqrt b\ne -\sqrt b$ (characteristic $\ne 2$), the left hand sides differ, hence so do the right hand soides, hence "$\pm$" is really "$-$". By adding and subtracting these equations we find that $a=m+n$ and $\sqrt b =2\sqrt{mn}$. Hence $m,n$ are roots of $0=X^2-(m+n)X+mn=X^2-a X+\frac b4$ and can be found as $$\frac{a\pm\sqrt{a^2-b}}2$$ More precisely:

• If $a^2-b$ is a square, this actually produces $m,n\in F$ with the property that $(\sqrt m +\sqrt n)^2=a+\sqrt b$, i.e. $\sqrt m + \sqrt n$ is a root of $X^2-(a+\sqrt b)$ as desired.
• If $a^2-b$ is not a square, no solutions for $m,n$ exist in $F$.

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 $$\sqrt{m}+\sqrt{n} =\frac{\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}}{2}+\frac{\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}}{2} = \sqrt{a+\sqrt{b}}.$$