It is known that $$ \sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}. $$ This formula is useful when $a^2-b$ is a perfect square: how can I know that this formula is unique? How can I know that it doesn't exist another way to reduce a nested radical such that $a^2-b$ is not a perfect square?
Wikipedia says that this is due to Galois theory, without proof. Is Galois theory necessary? Does it exist a "simpler" alternative argument?
It really depends on what you want to show. If you just want to find a condition on when you can write $\sqrt{a+\sqrt{b}}$ as $c\sqrt{d}+e\sqrt{f}$, then you can do that with simple algebra.
If you want to show that you can't write it as a sum of any number of non-nested radicals, then that gets more tricky, and I think you're going to have to get Galois theory involved.
First the simple algebra version
Suppose $\sqrt{a+\sqrt{b}}=c\sqrt{d}+e\sqrt{f}$ for some rational numbers $c,d,e,f$. Then $a+\sqrt{b} = c^2d+e^2f+2ce\sqrt{df}$, so (assuming $b$ isn't a perfect square) $df = b$ and $2ce=1$, and we can substitute $f=\frac{b}{d}$ and $e=\frac{1}{2c}$ in. Then $$a=c^2d+\frac{b}{4c^2d}.$$ Let $x=c^2d$ and multiply through by $x$, giving $ax=x^2+\frac{b}{4}$. Then there is a solution to this equation for $x$ in the rationals if and only if the discriminant, $a^2-4\frac{b}{4}= a^2-b$ is a perfect square.
Galois theory version
Let's first consider the general case, where, in particular, $a^2-b$ is not a square. Then (generically) the minimal polynomial of $\sqrt{a+\sqrt{b}}$ is $y^4-2ay^2+a^2-b$, which has the four roots $\pm\sqrt{a\pm\sqrt{b}}$. You can check that this will genuinely be irreducible as long as (1) $b$ is not a square in the rationals and (2) there do not exist $e,f\in\mathbb{Q}$ with $f^2=a^2-b$ and $e^2=2(f+a)$. (If such an $e$ and $f$ do exist, then it factors as $(y^2+ey+f)(y^2-ey+f)$).
Now the the splitting field of this polynomial, $K$, has degree 8 in general, and the splitting field can be given the equivalent presentation $\mathbb{Q}[\sqrt{b},\sqrt{a+\sqrt{b}},\sqrt{a-\sqrt{b}}]$, where each added generator is quadratic (in general) over the previous generators.
Let $G=\operatorname{Gal}{K/\mathbb{Q}}$. $G$ has order 8, and we can work out what it is. Since $K$ contains $\mathbb{Q}[\sqrt{b}]$, which is a normal extension of $\mathbb{Q}$, $G$ has a normal subgroup $N$ of order four. Morever, if $\phi$ is an automorphism of $K$ not in $N$, then $\phi(\sqrt{b})=-\sqrt{b}$, since $G/N\cong \operatorname{Gal}(\mathbb{Q}[\sqrt{b}]/\mathbb{Q})$. Then if we consider the polynomial $x^2-a-\sqrt{b}$, whose roots are $\pm\sqrt{a+\sqrt{b}}$, applying $\phi$ we get $\phi(x)^2-a+\sqrt{b}$, so $\phi(\pm\sqrt{a+\sqrt{b}})$ is a root of $x^2-a+\sqrt{b}$, so it must be one of $\pm\sqrt{a-\sqrt{b}}$. In other words $\phi$ doesn't fix the subfields $\mathbb{Q}[\sqrt{a+\sqrt{b}}]$ or $\mathbb{Q}[\sqrt{a-\sqrt{b}}]$. So those subfields are not normal. Since $G$ has non-normal subgroups, it isn't abelian.
Now all we need for later is that $G$ isn't abelian, but we can also identify the isomorphism class of $G$ without much additional work. Since it has order 8 it is either the quaternion group of order 8 or the dihedral group of order 8. But the quaternion group only has one subgroup of order 2, so in fact $G$ must be the dihedral group of order 8. Now it's also doable to work out what the automorphisms are and the isomorphism with the dihedral group, but that's not necessary here.
Our goal is to prove that we can't write $\sqrt{a+\sqrt{b}}$ in the form $$\alpha_0 + \sum_{i=1}^n \alpha_i \sqrt{\beta_i},$$ for $\alpha_i,\beta_i\in\mathbb{Q}$. If we could, then by considering the field $L=\mathbb{Q}[\beta_1,\ldots,\beta_n]$, which is normal, since it is the compositum of the normal field extensions $\mathbb{Q}[\beta_i]$, so if $y^4-2ay^2+a^2-b$ has a root in $L$, it factors completely in $L$, which means that $L$ contains $K$. However the Galois group of $L/\mathbb{Q}$ is $C_2^m$ for some integer $m$, which is abelian. However if $K\subseteq L$, then $G$ is a quotient of $C_2^m$, which would mean that $G$ would have to be abelian as well, but we already know that that isn't the case. Contradiction.
The non-generic cases:
What are the non-generic cases? Well, it's where our field $K$ doesn't have the full degree of 8 over $\mathbb{Q}$, or in other words when one of the generators is already contained in the field we're adding it to. Certainly if $b$ is a square in $\mathbb{Q}$, then $\sqrt{a+\sqrt{b}}$ is already expressed as a sum of square roots of elements of $\mathbb{Q}$. Next, we can consider the case where $a+\sqrt{b}$ is already a square in $\mathbb{Q}[\sqrt{b}]$, however, in that case, it will turn out that if $a+\sqrt{b} = (c+d\sqrt{b})^2$, then $a-\sqrt{b} = (c-d\sqrt{b})^2$, so this is actually a more specific case of the last case, which is when $a-\sqrt{b}$ is already a square in $\mathbb{Q}[\sqrt{a+\sqrt{b}}]$.
Now it'll turn out that the condition that $a-\sqrt{b}$ is a square in $\mathbb{Q}[\sqrt{a+\sqrt{b}}]$ is true if and only if $a^2-b$ is a square in $\mathbb{Q}[\sqrt{b}]$, if and only if there is a rational number $c$ such that $a^2-b =c^2$ or $a^2-b=bc^2$.
Proof:
First, if $a^2-b = (c+d\sqrt{b})^2$, then $a^2-b = c^2+2cd\sqrt{b}+bd^2$, and since $b$ isn't a square, this implies that either $c=0$ or $d=0$. If $d=0$, then $a^2-b=c^2$, and if $c=0$, then $a^2-b=bd^2$. So the final condition is equivalent to the second one.
Now if $a^2-b =c^2$, for $c\in \mathbb{Q}[\sqrt{b}]$, then consider $$\left(\frac{(a-\sqrt{b})\sqrt{a+\sqrt{b}}}{c}\right)^2 = \frac{(a^2+b-2a\sqrt{b})(a+\sqrt{b})}{a^2-b}$$ $$=\frac{a^3+a^2\sqrt{b}+ab+b\sqrt{b}-2a^2\sqrt{b}-2ab}{a^2-b} =\frac{a^3-a^2\sqrt{b}-ab+b\sqrt{b}}{a^2-b}$$ $$=\frac{(a^2-b)(a-\sqrt{b})}{a^2-b} = a-\sqrt{b}.$$ So $a-\sqrt{b}$ is a square in $\mathbb{Q}[\sqrt{a+\sqrt{b}}]$.
Now suppose that we knew that $a-\sqrt{b}$ was a square in $\mathbb{Q}[\sqrt{a+\sqrt{b}}]$. Let $u=\sqrt{a+\sqrt{b}}$ for notational simplicity. Then $K=\mathbb{Q}[\sqrt{a+\sqrt{b}}]$ is a splitting field, and therefore Galois over $\mathbb{Q}$. Let $v\in K$ be such that $v^2=a-\sqrt{b}$. Then $(uv)^2 = a^2-b$. Now if the degree of the minimal polynomial for $u$ over $\mathbb{Q}$ were 2, then we'd be in the case where $u\in\mathbb{Q}[\sqrt{b}]$, in which case $v$ is also in $\mathbb{Q}[\sqrt{b}]$, so $uv\in \mathbb{Q}[\sqrt{b}]$, and we are done. Otherwise, the minimal polynomial has degree four, which tells us that $G$ acts transitively on the roots of the minimal polynomial. Then the elements of $G$ can be identified with where they send $u$. Then consider $\phi_{-u}$, the automorphism that sends $u$ to $-u$. $\sqrt{b}=u^2-a$, so $\phi_{-u}(\sqrt{b}) = (-u)^2-a=\sqrt{b}$. Therefore $\phi_{-u}$ generates the subgroup corresponding to the subfield $\mathbb{Q}[\sqrt{b}]$. Now $\phi_{-u}(c)$ must be one of $\pm c$ because $\phi_{-u}(u)=-u$ and $\phi_{-u}(-u)=u$, but if $\phi_{-u}(v)=v$, then we'd have that $v\in \mathbb{Q}[\sqrt{b}]$, which would tell us that $u\in\mathbb{Q}[\sqrt{b}]$, which we are assuming isn't the case. So $\phi_{-u}(v)=-v$. Therefore $\phi_{-u}(uv)=(-u)(-v) = uv$. So $uv\in\mathbb{Q}[\sqrt{b}]$.
That completes the proof of our lemma. But it turned out that there were two cases, where either $a^2-b=c^2$ or $a^2-b=bc^2$. Let's consider the case where $a^2-b=bc^2$. Then $uv = c\sqrt{b}$, so we can work out what $\phi_v$ and $\phi_{-v}$ are. $\phi_v(uv)=-c\sqrt{b}=-uv=v\phi_v(v)$, so $\phi_v(v)=-u$. . Then $\phi_v(u)=v$, and $\phi_v^{2}(u)=-u$, so in particular $\phi_v^2\ne 1$. So $\phi_v$ has order four. That means that the Galois group in this case is $C_4$, which again can't show up as a quotient of $C_2^m$, which means that we can't write $\sqrt{a+\sqrt{b}}$ as a sum of non-nested radicals.
So the only possible case where we can rewrite $\sqrt{a+\sqrt{b}}$ as a sum of non-nested radicals (assuming $b$ is not itself a square) is when $a^2-b$ is a square in the rationals, in which case we already know how to do it, so this is an if and only if.
