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I want to understand how can the code below shows that $\left(\frac{\frac{-1+\sqrt{5}}{2}+\sqrt{\frac{-\sqrt{5}-5}{2}}}{2}\right)^5=1$

{
V=rnfequation(T^2-5,x^2-Mod((-T-5)/2,T^2-5),1);
a=V[2];
b=x-V[3]*a;
lift(((a-1+2*b)/4)^5)
}

recall :
rnfequation(nf,pol,{flag = 0})
given a number field nf as output by nfinit (or simply a polynomial) and a polynomial pol with coefficients in nf defining a relative extension L of nf, computes the absolute equation of L over Q.

If flag is non-zero, outputs a 3-component row vector [z,a,k], where z is the absolute equation of L over Q, as in the default behaviour, a expresses as an element of L a root alpha of the polynomial defining the base field nf, and k is a small integer such that theta = beta+kalpha where theta is a root of z and beta a root of pol.

Thanav
  • 129

1 Answers1

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By using the property that each side of a regular decagon inscribed in a circumference is the golden part of the radius, we get that

$\cfrac{\text{side}}{\text{radius}}=\cfrac{-1+\sqrt{5}}{2}\;$ and $\;\sin 18^\circ=\cfrac{\text{side / 2}}{\text{radius}}=\cfrac{-1+\sqrt{5}}{4}\;.$

Moreover,

$\cos 18^\circ=\sqrt{1-\sin^218^\circ}=\sqrt{1-\left(\cfrac{-1+\sqrt{5}}{4}\right)^2}=\\=\sqrt{\cfrac{5+\sqrt{5}}{8}}=\cfrac{1}{2}\sqrt{\cfrac{\sqrt{5}+5}{2}}\;.$

Since $\;\cfrac{\cfrac{-1+\sqrt{5}}{2}+\sqrt{\cfrac{-\sqrt{5}-5}{2}}}{2}=\sin 18^\circ+i\cos 18^\circ\;,\;$ then, by applying De Moivre’s formula, we obtain,

$\left(\cfrac{\cfrac{-1+\sqrt{5}}{2}+\sqrt{\cfrac{-\sqrt{5}-5}{2}}}{2}\right)^5=\left(\sin 18^\circ+i\cos 18^\circ\right)^5=\\=\sin 90^\circ+i\cos 90^\circ=1\;.$

Angelo
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    Aside from the fact that this doesn't really answer the question, I'm slightly concerned that this answer could be circular, at least as you've written it. How did you prove that $\sin(\pi/10) = \frac{-1 + \sqrt{5}}4$? – Mathmo123 Aug 30 '20 at 11:45
  • @Mathmo123, look at my answer now. – Angelo Aug 30 '20 at 13:19