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I was looking for the exact solutions of $\cos\frac{2\pi}{257}$, it lead me to the following expressions.

$A=-8\sqrt{514+18\sqrt{257}+12\sqrt{X}+4\sqrt{12593+561\sqrt{257}-98\sqrt{X}-400\sqrt{Y}}}-4\sqrt{514+18\sqrt{257}+12\sqrt{X}-4\sqrt{12593+561\sqrt{257}-98\sqrt{X}-400\sqrt{Y}}}-4\sqrt{514-18\sqrt{257}-12\sqrt{Y}+4\sqrt{12593-561\sqrt{257}-400\sqrt{X}+98\sqrt{Y}}}-4\sqrt{514+18\sqrt{257}-12\sqrt{X}+4\sqrt{12593+561\sqrt{257}+98\sqrt{X}+400\sqrt{Y}}}+\frac{1}{4}*[15+\sqrt{257}+2\sqrt{Y}+2\sqrt{257+15\sqrt{257}+16\sqrt{X}+14\sqrt{Y}}]*\sqrt{514-18\sqrt{257}+12\sqrt{Y}+4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}$.

with $X=\frac{257+\sqrt{257}}{2}$, $Y=\frac{257-\sqrt{257}}{2}$

Using numerical computation, I noticed that :

$A=6\sqrt{514-18\sqrt{257}+12\sqrt{Y}+4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}$

I did not succeed in doing the demonstration. Do you think it would be possible to demonstrate such a result?

Thank you and good luck.

user178256
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  • Can you translate in English? – taninamdar Oct 02 '14 at 14:39
  • How did you get this monster ? By the way, being French myself, I think it could be good (even for you) that you write in English. Amitiés :-) – Claude Leibovici Oct 02 '14 at 14:39
  • Also, did you mean $X, Y = \frac{257\pm\sqrt{257}}{2}$ or $X, Y = \frac{257 \pm 57\sqrt2}{2}$ at the last line? – Yiyuan Lee Oct 02 '14 at 14:52
  • @YiyuanLee. I bet it is the first one. I shall wait for your answer about this monster ! Cheers :-) – Claude Leibovici Oct 02 '14 at 14:54
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    @ClaudeLeibovici I suspect that it may actually be a very close approximation derived through some numerical method. A $4$ lines long equation is just crazy! – Yiyuan Lee Oct 02 '14 at 15:00
  • @YiyuanLee. I agree, for sure ! But may be, using $\cos(2x)=2\cos^2(x)-1$ could help you . Just a joke ! I really wonder how the OP got that; it is not an approximation, I guess. – Claude Leibovici Oct 02 '14 at 15:04
  • so in the french version you said you couldn't prove either of them nor prove that the two expressions are equal ? for which equality are you needing help ? – mercio Oct 02 '14 at 15:20
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    It seems that $257$ is a Fermat Prime. One can construct a $257$-sided regular polygon to compute the exact cosine of $\frac{2\pi}{257}$, similar to the approach of computing the cosines of $\frac{2\pi}{5}, \frac{2\pi}{17}, \frac{2\pi}{65537}$. Apparently the same question has already been asked here http://math.stackexchange.com/questions/516142/how-does-cos2-pi-257-look-like-in-real-radicals – Yiyuan Lee Oct 02 '14 at 15:23
  • does anyone else evaluate the second expression to $158.3484\ldots$ ? – mercio Oct 02 '14 at 17:00
  • @LeeYiyuan: He was close. – Tito Piezas III Dec 18 '14 at 22:05
  • @mercio: Yes, but he needs a similar expression to get to $\zeta_{257}$. – Tito Piezas III Dec 18 '14 at 22:07

1 Answers1

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Close, but not quite. As you defined, let,

$$X =\tfrac{257+\sqrt{257}}{2},\quad\quad Y =\tfrac{257-\sqrt{257}}{2}$$

with your conjugates,

$$A=6\sqrt{514-18\sqrt{257}+12\sqrt{Y}\color{red}{+}4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}=158.3484\dots$$

$$B=6\sqrt{514-18\sqrt{257}+12\sqrt{Y}\color{red}{-}4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}=25.3189 \dots$$

Given the $257$th root of unity $\zeta_{257} = e^{2\pi\,i/257}$. Then,

$$z=\sum_{k=1}^{64} {\zeta_{257}}^{22^k} =-22+\frac{A^2+B^2+12^3\,\sqrt{257}+24^2}{12^3}=9.246073\dots$$

In fact, $z$ is a root of,

$$z^4 + z^3 - 96z^2 - 16z + 256 = 0$$

Tito Piezas III
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