$\sqrt{n+\sqrt{2}}\inQ[e^{i\pi/8}]$的$n$

hbghlyj

$\frac{1}{\sqrt{2}}\left((1+\sqrt{2})^{2 n-1}-(1-\sqrt{2})^{2 n-1}\right)=$2, 10, 58, 338, 1970, 11482…
$\sqrt{2-\sqrt{2}}=2 \sin\left(\frac{\pi }{8}\right),\quad\sqrt{2+\sqrt{2}}=2 \cos \left(\frac{\pi }{8}\right)$
$\sqrt{10+\sqrt{2}}=2 \left(2 \sin \left(\frac{\pi }{8}\right)+\cos \left(\frac{\pi }{8}\right)\right)$ WA
$\sqrt{58-\sqrt2}=2 \left(5 \sin \left(\frac{\pi }{8}\right)+2 \cos \left(\frac{\pi }{8}\right)\right)$ WA
$\sqrt{338+\sqrt{2}}=2\left(12 \sin \left(\frac{\pi}{8}\right)+5 \cos \left(\frac{\pi}{8}\right)\right)$ WA
$\sqrt{1970-\sqrt{2}}=2\left(29 \sin \left(\frac{\pi}{8}\right)+12 \cos \left(\frac{\pi}{8}\right)\right)$ WA
如何证明呢？
也发到math.stackexchange.com/questions/4906578/问问。
自己发了一个直接计算的证明。
或许有更好的证明。

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相关帖子 $\sqrt{13+2\sqrt{13}}$属于$\Bbb Q[e^{i\pi/n}]$的最小$n$是26：\begin{aligned}&\sqrt{13+2\sqrt{13}}=2 \sin \left(\frac{\pi }{13}\right)+2 \sin \left(\frac{2 \pi }{13}\right)+2 \sin \left(\frac{3 \pi }{13}\right)+2 \cos \left(\frac{\pi }{26}\right)-2 \cos \left(\frac{3 \pi }{26}\right)+2 \cos \left(\frac{5 \pi }{26}\right)\\&=2\left(\cos \left(\frac{\pi}{52}\right)-\sin \left(\frac{\pi}{52}\right)\right)\left(\sin \left(\frac{\pi}{52}\right)+\cos \left(\frac{\pi}{52}\right)\right)\left(3+2 \sin \left(\frac{\pi}{26}\right)+2 \sin \left(\frac{5 \pi}{26}\right)-4 \cos\left(\frac{\pi}{13}\right)+2 \cos \left(\frac{2 \pi}{13}\right)\right)
\end{aligned}

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