The Exercise states:
Let $F$ be a field of characteristic $\neq 2$. Let $a$, $b$, 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 some $m$ and $n$ in $F$ is that $a^2 - b$ is a square in $F$. Use this to determine when the field $\mathbb{Q}(\sqrt{a + \sqrt{b}})$ ($a, b \in \mathbb{Q}$) is biquadratic over $\mathbb{Q}$.
I know a question has been asked pertaining to this exercise before on here, but the part that I'm stuck on has never been answered to, which is why I'm reposting it. I've already been able to prove the first part of the problem.
My conjecture is that
A necessary and sufficient condition for $\mathbb{Q}(\sqrt{a + \sqrt{b}})$ being biquadratic (i.e. equal to $\mathbb{Q}(\sqrt{n}, \sqrt{m})$ for some $n, m, nm$ not squares) is that $a^2 - b$ is a square and $b$ is not a square.
And it's easy for me to prove that if $a^2 - b$ is a square and $b$ is not a square then the extension is biquadratic.
Edit: Apperantely this is enough for the sufficient condition, but I need something extra on the necessary to avoid the cases where $\sqrt{n}$ and $\sqrt{m}$ are rational. Woops.
However, the converse is what's giving me trouble. I'd show my work, but suffice to say that it's a huge mess that's going nowhere. Mainly because I've tried things that are too convoluted like for instance setting $\sqrt{a + \sqrt{b}}$ equal to a linear combination of $1, \sqrt{n}, \sqrt{m}, \sqrt{mn}$, and squaring all of that.
If there's any clean method I would appreciate it, because the only thing that has occurred to me is doing as above, then setting $\sqrt{n}, \sqrt{m}, \sqrt{mn}$ equal to linear combinations of the basis elements of $\mathbb{Q}(\sqrt{a + \sqrt{b}})$, then finally solving for $a$ and $b$; and that would be incredibly messy.
Well actually, one other avenue that I've considered is assuming that $a = k(m + n)$ for some rational $k$, and then proving that this implies $\sqrt{b} = 2k\sqrt{mn}$. This would eventually imply $a^2 - b = k^2(m - n)^2$.
One last thing that I've noticed is that if the extension is biquadratic, then we know $(x^2 - a)^2 - b$ is the minimal polynomial of $\sqrt{a - \sqrt{b}}$ (since it's of degree 4). So if I could somehow show that $\sqrt{a - \sqrt{b}}$ is a root of $(x^2 - (m + n))^2 - 4mn$ (the minimal polynomial of $\sqrt{m} + \sqrt{n}$), preferably without directly demonstrating that $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n}$ beforehand then the constant coefficients ($a^2 - b$ and $(m - n)^2$) would be equal.
Ok, so I figured out how to prove this by showing that $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n}$ for some nonsquare $m$ and $n$. I do still wonder if there's a slicker way by demonstrating directly that $a^2 - b$ is square though.
Assume $\sqrt{b} = p_1 + p_2\sqrt{n} + p_3\sqrt{m} + p_4 \sqrt{mn}$. Squaring, we get that the coefficients of all the radial terms must be $0$. Thus $p_3p_4n = -p_1p_2$, $p_1p_4 = -p_2p_3$, and $p_1p_3 = -p_2p_4m$. If $p_1 \neq 0$, then $p_1^2 p_3 = p_2^2p_3m$. If $p_3 \neq 0$, then we get that $m = (\frac{p_1}{p_2})^2$, a contradiction of the assumption that it is not square. If $p_3 = 0$, then we have that $p_1p_2 = 0$, $p_1p_4 = 0$, and $p_2p_4 =0$. Therefore, it would be the case that only $p_1$ is nonzero, contradicting the fact that $b$ is nonsquare.
Thus $p_1 = 0$,To satisfy the equations, then two of $p_2, p_3$, and $p_4$ must be zero. WLOG (the other two cases work out in a very similar manner), assume $p_2$ is the nonzero element.
Now assume $\sqrt{a + \sqrt{b}} = q_1 + q_2\sqrt{m} + q_3 \sqrt{n} + q_4 \sqrt{mn}$ for some rational values $q_i$. Then $$a + \sqrt{b} = q_1^2 + q_2^2m + q_3^2n + q_4^2mn + (2q_2q_3 + 2q_1q_4)\sqrt{mn} + 2(q_1q_2 + q_3q_4n)\sqrt{m} + 2(q_1q_3 + q_2q_4)\sqrt{n}.$$
Again, since the terms $1, \sqrt{m}, \sqrt{n}$ and $\sqrt{mn}$ are linearly independent, we derive the equations
$$q_1q_2 + q_3 q_4n = p_2.$$ $$q_1 q_3 = -q_2q_4m.$$ $$q_1q_4 = -q_2q_3.$$
Suppose $q_1 \neq 0$. Then if $q_3 \neq 0$, it is the case that $q_1^2 = q_2^2m$, a contradiction again of the fact that $m$ is not square. However, if $q_3$ is equal to $0$, then $q_2q_4 = 0$, $q_1q_4 = 0$. For $p_2$ to be nonzero and these two equations to be satisfied, then $q_4$ must be equal to $0$ also. But this would make $\sqrt{a + \sqrt{b}}$ a degree $2$ element with respect to $\mathbb{Q}$. Whereas since the extension is biquadratic, then by definition it must be of degree $4$. Thus, $q_1 = 0$.
Then $q_2 = 0$ also, and $q_3, q_4$ are nonzero elements. Therefore, $\sqrt{a + \sqrt{b}} = \sqrt{q_3^2n}+ \sqrt{q_4^2mn}$, as desired.