\documentclass{article}
\usepackage{amsmath}
\begin{document}
\title{Solving $x^5-x^4-1=0$}
\author{Amit Yaron}
\date{Jul 9, 2024}
\maketitle
Don't use the Abel-Ruffini theorem as an excuse to skip that equation.
Find a trick to factorize it into a cubic and a quadratic polynomial.
5 and 4 are two consecutive numbers, and none of them is divisible by 3.
And, there are the two identities:
\begin{equation}x^3-1=(x-1)(x^2+x+1)\end{equation}
and
\begin{equation}x^3+1=(x+1)(x^2-x+1)\end{equation}
which means that if $x$ is a complex number:\\
$x^2+x+1=0\Rightarrow x^3=1$\\
and
$x^2-x+1=0\Rightarrow x^3=-1$\\
\\
Obe conclusion is that if $m,n$ are non-negative integers, then:\\
$x^2+x+1\mid x^{3m+2}+x^{3n+1}+1$\\
The proof is simple:\\
$x^2+x+1=0\Rightarrow x^3=1\Rightarrow x^{3m+2}+x^{3n+1}+1=x^2+x+1$\\
\\
Finding polynomials divisible by $x^2-x+1$ is more complicated. Let
us just plug $x^3=-1$ into our equation:\\
$x^2-x+1=0\Rightarrow\\
\Rightarrow x^3=-1\Rightarrow\\
\Rightarrow x^5-x^4-1=-x^2+x-1=-(x^2-x+1)$\\
\\
Now, let us solve the equation:
\begin{align*}&x^5-x^4-1=0\Rightarrow\\
\Rightarrow&x^3(x^2-x)-1=0\Rightarrow\\
\Rightarrow&x^3(x^2-x+1)-x^3-1=0\Rightarrow\\
\Rightarrow&x^3(x^2-x+1)-(x^3+1)=0\Rightarrow\\
\Rightarrow&x^3(x^2-x+1)-(x+1)(x^2-x+1)=0\Rightarrow\\
\Rightarrow&(x^3-x-1)(x^2-x+1)=0\end{align*}
The solution of the quadratic equation is:
\begin{equation}x=\frac{1\pm\sqrt3i}2\end{equation}
The cubic equation will be solve using Cardano-Tartaglia formula:\\
Find $u,v$ such that:
$x=u+v\Rightarrow x^3-x-1=0$\\
From the formula for cubing a sum of 2 numbers:
\begin{align}&(u+v)^3=u^3+v^3+3uv(u+v)\nonumber\Rightarrow\\
\Rightarrow& x^3=3uvx+u^3+v^3\end{align}
Equating coefficients of (4) with those of $x^3=x+1$, we get that:\\
$\begin{cases}3uv=1\\u^3+v^3=1\end{cases}\Rightarrow\\
\Rightarrow \begin{cases}27u^3v^3=1\\u^3+v^3=1\end{cases}$\\
Let us plug $v^3=1-u^3$ into the first equation:
\begin{align*}&27u^3(1-u^3)=1\Rightarrow\\
\Rightarrow&27u^3-27u^6=1\Rightarrow\\
\Rightarrow&27u^6-27u^3+1=0\Rightarrow\\
\Rightarrow& u^3=\frac{27\pm\sqrt{27^2-108}}{54}\Rightarrow\\
\Rightarrow& u^3=\frac{27\pm\sqrt{621}}{54}\Rightarrow\\
\Rightarrow& u^3=\frac{27\pm3\sqrt{69}}{54}\Rightarrow\\
\Rightarrow&u^3=\frac{9\pm\sqrt{69}}{18}\Rightarrow\\
\Rightarrow&u^3=\frac12\pm\sqrt{\frac{69}{324}}\Rightarrow\\
\Rightarrow&u^3=\frac12\pm\sqrt{\frac{23}{108}}\end{align*}
Now, WLOG, let us take:
$\begin{cases}u^3=\frac12+\sqrt{\frac{23}{108}}\\
v^3=\frac12-\sqrt{\frac{23}{108}}\end{cases}$\\
Let us take cube roots, then:
$\begin{cases}u=\sqrt[3]{\frac12+\sqrt{\frac{23}{108}}}\\
v=\sqrt[3]{\frac12-\sqrt{\frac{23}{108}}}\end{cases}$\\
Now, the first root of the cubic equation will be\\
$x=u+v\approx1.324717957244746$\\
For more solution, let us define:\\
$w=\frac{-1+\sqrt{3}i}2$\\
The second solution will be:\\
$x=uw+vw^{*}\approx-0.662358978622373+0.5622795120623011i$\\
The third solution will be:\\
$x=uw^{*}+vw\approx-0.662358978622373-0.5622795120623011i$
