Can you show that the regular 15 sided polygon is constructible

Question

It is known from basic geometry that the regular 15 sided polygon is constructible. It can be shown that Cos2π/15 radian is a root of the equation: x7 +1/2 X6 - 3/2 X5 - 5/8 X4 + 5/8 X3 + 3/16 X2 - X/16 - 1/128 = 0 where X= Cos 24°. Can you demonstrate this? This equation is algebraically solvable in terms of square roots, because this polygon is constructible. Can you show that one form of the solution is the following: x= Cos 24°= (-1+√w)/4 , Where w = 11+2√y , and y = 3(5+2√5).

Answer 1

OK, i will type that solution. Let us define first: $$ \begin{aligned} \xi &= \cos 24^\circ + i\sin 24^\circ=\exp\frac{2\pi i}{15}\ ,\\ c &=2\cos\frac{2\pi i}{15}=\xi+\frac 1\xi\ ,\\ x&=\frac c2=\cos 24^\circ=\cos \frac{2\pi}{15}\ ,\\ y &= 3(5+2\sqrt 5)>0\ ,\\ w &= 11+2\sqrt y\ .\\[3mm] &\qquad\text{Then we have:}\\[3mm] \xi&\text{ is a root of the cyclotomic polynomial $\Phi_{15}$,}\\ \Phi_{15}(X) &=\frac{X^{15}-1}{\Phi_5(X)\Phi_3(X)\Phi_1(X)}\\ &=\frac{X^{15}-1}{\frac{X^5-1}{X-1}\cdot\Phi_3(X)\cdot(X-1)}\\ &=\frac{X^{10}+X^5+1}{X^2+X+1}\\ &=X^8 - X^7 + X^5 - X^4 + X^3 - X + 1\ .\\[3mm] &\qquad\text{It is real and reciprocal. So we have:}\\[3mm] 0&=\Phi_{15}(\xi)=\Phi_{15}(\bar\xi)=\Phi_{15}\left(\frac 1\xi\right)\ . \\[3mm] &\qquad\text{We will show $c^4-c^3-4c^2+4c+1=0$:}\\[3mm] \xi^4&(c^4-c^3-4c^2+4c+1)\\ &=(\xi^2+1)^4-\xi(\xi^2+1)^3-4\xi^2(\xi^2+1)^2 +4\xi^3(\xi^2+1)+\xi^4\\ &=(\xi^8+4\xi^6+6\xi^4+4\xi^2+1) \\ &\qquad-(\xi^7+3\xi^5+3\xi^3+1) \\ &\qquad\qquad-4(\xi^6+2\xi^4+\xi^2) \\ &\qquad\qquad\qquad+4(\xi^5+\xi^3) \\ &\qquad\qquad\qquad\qquad+\xi^4 \\ &=\Phi_{15}(\xi) \\ &=0\ . \end{aligned} $$

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The above (minimal) equation for $c=2\cos(2\pi/15)$ implies an obvious equation for $x=c/2= \cos(2\pi/15)$, which is: $$ x^4 - \frac 12x^3 - x^2 + \frac 12x + \frac 1{16}=0\ . $$ The polynomial in the OP has the polynomial above as a factor, in fact explicitly: $$ \left( x^{7} + \frac{1}{2} \, x^{6} - \frac{3}{2} \, x^{5} - \frac{5}{8} \, x^{4} + \frac{5}{8} \, x^{3} + \frac{3}{16} \, x^{2} - \frac{1}{16} \, x - \frac{1}{128} \right) \\ = \left(x^4 - \frac 12 x^3 - x^2 + \frac 12 x + \frac1{16}\right) \left(x^2 + \frac 12 x - \frac 14\right) \left(x + \frac 12\right) \ . $$ This clears the first question in the OP.

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The second question, regarding the expression using radicals for $x$. Let us show: $$ \begin{aligned} (*)\qquad x &=\frac 18\Big(\ 1+\sqrt 5+\sqrt{6(5-\sqrt 5)} \ \Big) \\ &= -\frac 14+ \frac 14\sqrt{11+2\sqrt{3(5+2\sqrt 5)}} =\frac 14(-1+\sqrt w) \ . \end{aligned} $$

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For the first equality, marked $(*)$ above, we easily identify the Galois conjugates, which are $$ x_{s,t}= \frac 18\Big(\ 1+s\sqrt 5+t\sqrt{6(5-s\sqrt 5)} \ \Big)\ ,\ s,t\in\pm 1\ . $$ (It is a good structural way to write so the solution(s).) Then we compute $$ \begin{aligned} &64(X-x_{s,+}) (X-x_{s,-}) \\ &\qquad = (8X-8x_{s,+}) (8X-8x_{s,-}) \\ &\qquad = (8X-s\sqrt 5-1)^2 - 6(5-s\sqrt 5) \\ &\qquad = \Big((8X)^2+5+1-16X-30\Big) -2s\sqrt 5\Big((8X-1)-3\Big) \\ &\qquad = 64\left[\ \left(X^2-\frac 14X-\frac 38\right) \ -\ \frac 18s\sqrt 5(2X-1) \ \right]\ . \\[3mm] &(X-x_{+,+}) (X-x_{+,-}) (X-x_{-,+}) (X-x_{-,-}) \\ &\qquad= \prod_{s\in\pm1 } (X-x_{s,+}) (X-x_{s,-}) \\ &\qquad = \left(X^2-\frac 14X-\frac 38\right)^2 -\frac 5{64}(2X-1)^2 \\ &\qquad = X^{4} - \frac{1}{2} \, X^{3} - X^{2} + \frac{1}{2} \, X + \frac{1}{16} \ . \end{aligned} $$ The above computation shows that all values of $x_{s,t}$ are the solutions of the above equation of degree four. It remains to numerically identify them with $\cos(2k\pi/15)$ for $k$ among the values in $1,2,\dots14$ relatively prime to $15$, and taken modulo $\pm$, so $1,2,4,7$. It turns out that we have $x=\cos(2\pi/15)=x_{++}$.

Note that this first formulas for the four roots involve only a "second recursivity level" of taking radicals.

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Let us show also the second expression using radicals for $x=\cos(2\pi/15)$. For this, we want to show $$ \frac 18\Big(\ 3+\sqrt 5+\sqrt{6(5-\sqrt 5)} \ \Big) \ \overset{(!)}{=\!=} \ \frac 14\sqrt{11+2\sqrt{3(5+2\sqrt 5)}} \ . $$ It is a claimed equality between positve reals. Equivalently after squaring: $$ \Big(\ 3+\sqrt 5+\sqrt{6(5-\sqrt 5)} \ \Big)^2 \ \overset{(!)}{=\!=} \ 4\left(\ 11+2\sqrt{3(5+2\sqrt 5)}\ \right) \ . $$ The LHS may be computed as $$ 9+5+(30-6\sqrt 5)+6\sqrt 5+6\sqrt{6(5-\sqrt 5)} +2\sqrt 5\cdot \sqrt{6(5-\sqrt 5)}\ . $$ We have a cancellation of $6\sqrt 5$. Then the integers $9+5+30$ sum up to $44$, which is also a term on the RHS. Note that the big radical remained on the LHS can be factored, and $6+2\sqrt 5=(\sqrt 5+1)^2$. It remains to show, after a new squaring: $$ (\sqrt 5+1)^4\cdot 6\cdot\sqrt5(\sqrt 5-1) \ \overset{(!)}{=\!=} \ 8^2\cdot 3(5+2\sqrt 5)\ . $$ On the LHS we can group $(\sqrt 5+1)(\sqrt 5-1)=5-1=4$. We simplify then $6\cdot 4=8\cdot 3$. We compute $(\sqrt 5+1)^3 =(8\sqrt 5+16)=8(\sqrt 5+2)$, and are done.

This completes all the points.

Answer 2

more to the point, the primitive 15th roots of unity satisfy $$ \omega^8 - \omega^7 + \omega^5 - \omega^4 + \omega^3 - \omega + 1 = 0 $$ while $x =2 \cos \frac{2 \pi}{15},$ which can be written as one of the $x =\omega + \frac{1}{\omega},$ satisfies $$ x^4 - x^3 - 4 x^2 + 4 x + 1 = 0 $$