Find all integers $m$ and $n$ such that $n\cos\frac{\pi}{m}= \sqrt{8+\sqrt{32+\sqrt{768}}}$

Question

Find all integers $m$ and $n$ such that: $$n\cos\frac{\pi}{m}= \sqrt{8+\sqrt{32+\sqrt{768}}}$$

I tried to find the relation of $m$ and $n$, but it's difficult for me. I need help! Thanks!

Answer 1

By applying the formula $$\cos x=\sqrt{\frac{1+\cos(2x)}2}$$ three times, you can obtain $n\cos k=$

By direct comparison, $$n= 4$$ $$\cos 8k=\frac12\implies 8k=\pm \frac\pi3$$ Thus, $m=\pm24$.

To prove uniqueness, consider $$|m|\ge2\implies\frac1{|m|}\le\frac12\implies \frac\pi{|m|}\le\frac\pi2$$

In the region, cosine is positive. Thus, $n>0$.

For a moment, assume $m>0$.

Since $n\cos x$ is strictly decreasing in the region $0\le x\le\pi/2$, it intersects a single value(a horizontal line) at most once. Thus $m=24, n=4$ is the unique solution for $m,n>0$.

Exploiting the evenness of cosine, $m=-24, n=4$ is also a solution.

ADDED:

One may question whether, with $a,b,c,d$ natural numbers and $b,d$ non-square numbers, the following is true: $$\sqrt{a+\sqrt{b}}=\sqrt{c+\sqrt{d}}\implies a=c, b=d$$

Suppose $a\ne c, b\ne d$, $a,b,c,d$ are natural numbers, $b,d$ non square numbers.

Manipulating $\sqrt{a+\sqrt{b}}=\sqrt{c+\sqrt{d}}$ by repeatedly squaring both sides and rearranging to get rid of square roots, one obtains

$$((a-c)^2-(b-d))^2=4(a-c)^2d$$

Note that the left hand side is a square number, thus so as the right hand side. Then, this requires either $d$ to be a square number, or $(a-c)=0$, which both oppose our supposition.

Thus, $$\sqrt{a+\sqrt{b}}=\sqrt{c+\sqrt{d}}\implies a=c, b=d$$ is true.

Similarly, the direct comparison I made is therefore justified.

p.s. Personally, I like @labbhattacharjee ‘s answer more than mine. His answer doesn’t require the ADDED part proof. Well, I think of half angle formula because of the square roots; his answer needs some math sense to see the hidden $\cos(60-45)$.

Answer 2

$$8+\sqrt{32+\sqrt{768}}=8+\sqrt{8(\sqrt3+1)^2}=8\left(1+\dfrac{\sqrt3+1}{2\sqrt2}\right)=8(1+\cos(60-45)^\circ)=16\cos^2(7.5^\circ)$$