Why does there seem to be a relation between the fifth roots of $32$ and $\varphi$ (the golden ratio)?

Question

I have the following problem on one of my assignments:

find all five fifth roots of 32.

What I find interesting is how $\varphi$ is related to the real component of most of the solutions. My intuition points me towards the definition of $\varphi$, namely $\varphi$ = $\frac{\sqrt{5} + 1}{2}$

Here are the solutions: $2$, $0.618+0.951i$, $-1.618+1.176i$, $-1.618-1.176i$, $0.618-1.902i$

Four of the solutions have real component (of what seems to be) either $-\varphi$ or $\varphi - 1$. It is not just a coincidence to do with rounding, it is exactly equal to $-\varphi$ and $\varphi - 1$ out to 10 decimal places.

Is there some deeper relation that I am missing? I was shocked to get this answer.

Answer 1

Note that

$$w^5=32e^{i2k\pi} \implies w_k= 2e^{i\frac{2k\pi}{5}}$$

with $k=0,1,2,3,4$ and for example

$$w_1 = 2\left(\cos \frac{2\pi}{5}+i \sin \frac{2\pi}{5}\right)$$

which is connected to golden ratio by the golden triangle

(image modified from credit 01, credit 02)

Answer 2

If $z^5=32$ then $$z=2 \exp\left(\frac {i2\pi k}5\right)=2 \left[\cos\left(\frac {2\pi k}5\right)+i\sin\left(\frac {2\pi k}5\right)\right]$$ for $k\in\{0,1,2,3,4\}$. The real part of $z$ is $$2\cos\left(\frac {2\pi k}5\right)=\exp\left(\frac {i2\pi k}5\right)+\exp\left(-\frac {i2\pi k}5\right)=x+\frac1x,$$ where $x^5=1$. When $x=1$, this equals $2$. Otherwise, factor $x^5-1$ as $$ \begin{aligned} x^5-1&=(x-1)(x^4+x^3+x^2+x+1)\\ &=(x-1)x^2\left(x^2+x+1+\frac1x+\frac1{x^2}\right)\\ &=(x-1)x^2\left[\left(x+\frac1x\right)^2+\left(x+\frac1x\right)-1\right]. \end{aligned} $$ So when $x\ne1$ we must have $\left(x+\frac1x\right)^2+\left(x+\frac1x\right)-1=0$, and the quadratic equation yields $x+\frac1x=\frac{-1\pm\sqrt 5}2$, which equals $-\phi$ or $\phi-1$.

Answer 3

Taking the factorization in @grand_chat's answer one step further, and using the conjugate notation $\,\overline\varphi = -\dfrac{1}{\varphi} = - \varphi+1\,$ for the second root of the golden ratio equation:

$$ x^5 - 1 \;=\; (x-1)\,\big(x^2 + \varphi x + 1\big)\,\big(x^2 + \overline\varphi x + 1\big) $$

This shows the presence of $\varphi$ in the factorization of $\,x^5 - 1\,$ over the reals, and confirms the connection with the $5^{th}$ roots of unity and associated trigonometric functions.

Perhaps more unexpectedly, this also facilitates an alternative proof of Fermat's Last Theorem in the special case $n=5$. Quoting from George Lowther's answer under Golden Number Theory:

Using the fact that $\mathbb{Z}[\varphi]$ is a unique factorization domain in which we can decompose $$ x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy) $$ we can give a proof of Fermat's Last Theorem for the case of exponent $5$.