Quadratic equation family with largest real root in Cyclotomic extension Let the $\alpha_{k}$ be the largest real root by absolute value of $ 2x^2-2kx-(k-1)=0$ for all $k\ge1$.
For what values of $k$ does $\alpha_{k}$ sit in a cyclotomic extension? How does one explicitly provide the extension when an $\alpha_{k}$ does sit in a cyclotomic extension?
 A: Here is a proof quadratic number fields lie in cyclotomic fields via Gauss sums.
Suppose $L/K/{\Bbb Q}={\Bbb Q}(\zeta_n)/{\Bbb Q}(\sqrt{d})/{\Bbb Q}$ is a tower of Galois extensions, where $d\in{\Bbb Z}$ is squarefree.
Then $K$ is the fixed field $L^H$ for some $H\subset G={\rm Gal}(L/{\Bbb Q})$ by Galois correspondence.
What naive assumption could we make to force $G$ to be as simple as possible? The simplest possible groups to work with are cyclic groups. In particular, say $n=p$ is prime $>2$ for convenience.
Furthermore $[K:{\Bbb Q}]=2=[G:H]$ by Galois correspondence. As $G$ is even-order cyclic, its unique index-two subgroup is the set of squares $H=G^2$.
An artificial way to create an $H$-invariant element of $L$ is to sum over the $H$-orbit of an $\alpha\in L$, or equivalently to apply the trace map ${\rm tr}_{L/K}$. Might as well apply it to $\zeta_p$ of all things:
$$h:=\sum_{k=1}^{p-1}\zeta_p^{k^2}=2\sum_{u\in H}\zeta_p^u.$$
Technically we have summed over elements of the $H$-orbit of $\zeta_p$ twice since $k^2=(-k)^2$, but this rescaling shouldn't affect $H$-invariance. The luckiest we could get is if $g=\sqrt{d}$ already. It can be checked that this is not quite true, so there is still more work to do. Now here's a magic trick:
$$0=\sum_{k=0}^{p-1}\zeta_p^k=1+\sum_{u\in H}\zeta_p^u+\sum_{v\in G\setminus H}\zeta_p^v\iff h=-1+\sum_{u\in H}\zeta_p^u-\sum_{v\in G\setminus H}\zeta_p^h.$$
The $1$ is aesthetically annoying, so let's get rid of it by setting $g=h+1$. Then we can express $g$ as
$$g:=\sum_{k=1}^{p-1}\left(\frac{k}{p}\right)\zeta_p^k$$
where the Legendre symbol is defined to take the values $+1,-1$ depending on whether $k$ is a quadratic residue (i.e. square mod $p$) or not respectively. Observe that the Legendre symbol is a group homomorphism $G\to\{\pm1\}$ with kernel $H$. This $g$ is called a Gauss sum. They come in much more advanced forms elsewhere in number theory. For a good resource see Hugh (pdf).
Finally, let's check if we're lucky and squaring $g$ yields something useful. In order to do this though we need to be clever and think of $g$ as a special case of a "discrete Fourier transform." Define
$$g_a:=\sum_{k=1}^{p-1}\left(\frac{k}{p}\right)\zeta_p^{ak}=\left(\frac{a}{p}\right)^{-1}\sum_{k=1}^{p-1}\left(\frac{ak}{p}\right)\zeta_p^{ak}=\left(\frac{a}{p}\right)g~~~(a\not\equiv0).$$
Now we compute an "inner product induced norm" in two different ways. First way:
$$\sum_{a=0}^{p-1}g_ag_{-a}=g_0^2+\sum_{a=1}^{p-1}\left(\frac{a}{p}\right)\left(\frac{-a}{p}\right)g^2=(p-1)\left(\frac{-1}{p}\right)g^2=(p-1)(-1)^{(p-1)/2}g^2.$$
This follows since $\left(\frac{a}{p}\right)\left(\frac{-a}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{a^2}{p}\right)=\left(\frac{-1}{p}\right)$ is constant wrt $a$ and $\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}$ by Legendre's formula $\left(\frac{a}{p}\right)\equiv a^{(p-1)/2}\bmod p$ and $g_0=0$. The second way of summing is
$$\sum_{a=0}^{p-1}g_ag_{-a}=\sum_{u=1}^{p-1}\sum_{v=1}^{p-1}\left(\frac{u}{p}\right)\left(\frac{v}{p}\right)\sum_{a=0}^{p-1}\zeta_p^{a(u-v)}=p\sum_{u=v=1}^n \left(\frac{uv}{p}\right)=(p-1)p$$
since the inner sum is $p$ if $u\ne v$ and $0$ otherwise. Equating results yields
$$(p-1)(-1)^{(p-1)/2}g^2=(p-1)p\iff g^2=(-1)^{(p-1)/2}p.$$
Unfortunately this only tells us that ${\Bbb Q}(\zeta_p)\supset{\Bbb Q}(\sqrt{(-1)^{(p-1)/2}p})\supset{\Bbb Q}$ for primes $p>2$. One can check though that $\sqrt{-1},\sqrt{2}\in{\Bbb Q}(\zeta_4)$, hence for any product $d=q_1\cdots q_m$ of distinct primes
$${\Bbb Q}(\sqrt{\pm d})\subseteq{\Bbb Q}(\zeta_4,\zeta_{q_{\large 1}},\cdots,\zeta_{q_{\large m}})\subseteq{\Bbb Q}(\zeta_{4d}).$$
