Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$

Question

Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.

I have tried it by substituting $x = \exp(2i\pi/5)$ but it is getting complicated.

Answer 1

Replacing the earlier sequence of hints with a solution now. The old aswer is scratched.

We know that $0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. As $x-1\neq0$ this implies that $$1+x+x^2+x^3+x^4=0.\qquad(*)$$ From $x^5=1$ we also deduce that $x^4=x^{-1}$. This allows a rewrite: $$ S=(x^2-x^3)(x^4-x)=(x^2-x^3)(x^{-1}-x)=(x-x^2)(1-x^2)=x-x^2-x^3+x^4. $$

Let's square this. We get $$ \begin{aligned} S^2&=(x^2+x^4+x^6+x^8)\\ &+2(-x^3-x^4+x^5+x^5-x^6-x^7).\qquad(**) \end{aligned} $$

Because $x^8=x^5\cdot x^3=x^3$ and $x^6=x$, the first term above is $$ (x^2+x^4+x^6+x^8)=x^2+x^4+x+x^3=-1, $$ by equation $(*)$. The latter term in parens is similarly simplified to $$ (-x^3-x^4+x^5+x^5-x^6-x^7)=2-(x^3+x^4+x+x^2)=3. $$ Plugging these both into $(**)$ gives $$ S^2=5. $$ So we know that $S=\pm\sqrt5$, and the remaining task is to determine the sign. From a picture of the unit circle in the complex plane, we see that all the terms in the r.h.s. of $$ S=x-x^2-x^3+x^4 $$ have positive real parts. Therefore $S=\sqrt5$.

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No trigonometry needed. Just $(*)$ and properties of roots of unity. Study Gauss sums for generalizations to primes $>5$.

Answer 2

Note $(x^2-x^3)(x^4-x)=-x^7+x^6+x^4-x^3$. The key is that $x$ is a root of unity, particularly one that gives $x^5=1$. This means $x^6=x$ and $x^7=x^2$ while $x^4=x^*$ and $x^3=(x^2)^*$ which means that our imaginary components fall out since $z+z^*=2a$ for $z=a+bi$. Squaring both sides to get rid of the radical reduces the problem nicely.

sorry for any errors as I've posted this from my phone.

Answer 3

As $x=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$

Using de Moivre's formula for positive integer $n$

$$x^n=\left(\cos\frac{2\pi}5+i\sin\frac{2\pi}5\right)^n=\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5$$

$$\implies x^5=\cos2\pi=1\text{ and }x^{-n}=\frac1{x^n}=\frac1{\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5}=\cos\frac{2n\pi}5-i\sin \frac{2n\pi}5$$

$$\implies x^n+x^{-n}=2\cos\frac{2n\pi}5$$

As $x^5=1,x^6=x,x^4=x^{-1},x^7=x^2,x^3=x^{-2}$

$$\implies (x^2-x^3)(x^4-x)=x^6+x^4-x^7-x^3=x+\frac1x-\left(x^2+\frac1{x^2}\right)$$ $$\implies (x^2-x^3)(x^4-x)=2\cos\frac{2\pi}5-2\cos\frac{4\pi}5=2\cos\frac{2\pi}5-2\cos(\pi-\frac\pi5)=2\cos\frac{2\pi}5+2\cos\frac\pi5>0\ \ \ \ (0)$$

Now if $y^5=1$

Using $n$th root of unity, $\displaystyle y=\cos\frac{2r\pi}5+i\sin \frac{2r\pi}5$ where $r=0,1,2,3,4$

$r=0\implies y=1\implies$ the roots of $\displaystyle \frac{y^5-1}{y-1}=0\iff y^4+y^3+y^2+y+1=0\ \ \ \ (1)$ are $\cos\frac{2r\pi}5+i\sin \frac{2r\pi}5$ where $r=1,2,3,4$

Observe that the last equation is Reciprocal Equation of the First type like this

So, divide either sides by $y^2,$ $$y^2+y+1+\frac1y+\frac1{y^2}=0 \implies \left(y+\frac1y\right)^2+\left(y+\frac1y\right)-1=0\ \ \ \ (2)$$

We have $y+\frac1y=2\cos\frac{2r\pi}5$

and as $\cos\frac{2(5-r)\pi}5=\cos(2\pi-\frac{2r\pi}5)=\cos \frac{2r\pi}5$

the roots of $(2)$ are $2\cos\frac{8\pi}5=\cos\frac{2\pi}5>0$ and $\cos\frac{6\pi}5=\cos(2\pi-\frac{2r\pi}5)=\cos\frac{4\pi}5=-\cos\frac\pi5<0$

Now, $2\cos\frac{2r\pi}5=y+\frac1y=\frac{-1\pm\sqrt5}2$ where $r=(1$ or $5-1=4)$ and $r=(2$ or $5-2=3)$

So, using Vieta's formulas, $2\cos\frac{2\pi}5+\left(-2\cos\frac\pi5\right)=-1$ and $2\cos\frac{2\pi}5\left(-2\cos\frac\pi5\right)=-1$

$$\implies 2\cos\frac{2\pi}5+2\cos\frac\pi5=\sqrt{\left(2\cos\frac{2\pi}5-2\cos\frac\pi5\right)^2+4\cdot2\cos\frac{2\pi}5\cdot2\cos\frac\pi5}=\sqrt{(-1)^2+4\cdot1}=\sqrt5$$

Answer 4

As was already mentioned, we have

$$0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$

$$|x|=|e^{2\pi i/5}|=1\implies x^2=x^{-3}\;,\;x=x^{-4}\;,\;x^{-k}=\overline{x^k}\implies$$

$$(x^2-x^3)^2(x^4-x)^2=(x^2-x^{-2})^2(x^{-1}-x)^2=(x-x^{-1})^4(x+x^{-1})^2=$$

$$=\left(2i\sin\frac{2\pi}5\right)^4\left(2\cos\frac{2\pi}5\right)^2=64\sin^4\frac{2\pi}5\cos^2\frac{2\pi}5=$$

$$=64\left(\frac14\sqrt{10+2\sqrt5}\right)^4\left(\frac14\left(-1+\sqrt5\right)\right)^2=5$$

Answer 5

$x^4+x^3+x^2+x+1=0$. Divide by $x^2$, and substitute $z=x+x^{-1}$, resulting in $z^2+z-1=0$. This has two solutions: $z_1=2cos72^\circ=\dfrac{-1+\sqrt5}{2}$, and $z_2=2cos144^\circ=\dfrac{-1-\sqrt5}{2}$.