\documentclass[a4paper]{article}
\usepackage{amsmath}
\begin{document}
\title{Find all Roots, Solve for Square First}
\author{Amit Yaron}
\date{Nov 16, 2025}
\maketitle
\section*{How to solve $x^7+7^7=(x+7)^7$}
Let us first write the equation as
\[
  (x+7)^7-x^7=7^7
\]
Here I'm going to use a substitution, that will make the odd powers of $x$
disappear.
Let us first mulitply both sides by $2^7$, so we'll work with integers.
\[
  2^7(x+7)^7-2^7x^7=2^77^7\Rightarrow (2x+14)^7-(2x)^7=14^7
\]
Now, let us substitute
\[
  t=2x+7
\]
Then
\[
  2x+14=t+7\text{ and } 2x=t-7
\]
And the equation will be
\[
  (t+7)^7-(t-7)^7=14^7
\]
Let us expand:
\begin{equation}
  (t+7)^7=t^7+7\cdot7t^6+21\cdot7^2t^5+35\cdot7^3t^4+35\cdot7^4t^3+21\cdot7^5t^2+7\cdot7^6t+7^7
\end{equation}
and
\begin{equation}
  (t+7)^7=t^7-7\cdot7t^6+21\cdot7^2t^5-35\cdot7^3t^4+35\cdot7^4t^3-21\cdot7^5t^2
+7\cdot7^6t-7^7
\end{equation}
Let us subtract (2) from (1) and get:
\[
  14\cdot7t^6+70\cdot7^3t^4+42\cdot7^5t^2+2\cdot7^7=14^7
\]
Now, two real values of x are roots of the original equation 
\[
  x=0\Rightarrow t=2x+7=7
\]
and
\[
  x=-7\Rightarrow t=2x+7=-7
\]
So $t^2=7^2$ is a solution. Let us start solving
\begin{align*}
  &14\cdot7t^6+70\cdot7^3t^4+42\cdot7^5t^2+2\cdot7^7-14^7=0\Rightarrow\\
  \Rightarrow &7t^6+5\cdot7^3t^4+3\cdot7^5t^2+7^6-14^6=0\Rightarrow\\
  \Rightarrow &7t^6+5\cdot7^3t^4+3\cdot7^5t^2+7^6-64\cdot7^6=0\Rightarrow\\
  \Rightarrow &7t^6+5\cdot7^3t^4+3\cdot7^5t^2-63\cdot7^6=0\Rightarrow\\
  \Rightarrow &7t^6+5\cdot7^3t^4+3\cdot7^5t^2-9\cdot7^7=0\Rightarrow\\
  \Rightarrow &7t^6-7^3t^4+6\cdot7^3t^4-6\cdot7^5t^2+9\cdot7^5t^2-9\cdot7^7=0\Rightarrow\\
  \Rightarrow &(t^2-49)(7t^4+6\cdot7^3t^2+9\cdot7^5)=0\Rightarrow\\
  \Rightarrow &(t^2-49)(t^4+6\cdot7^2t^2+9\cdot7^4)=0\Rightarrow\\
  \Rightarrow &(t^2-49)\big({(t^2)}^2+2\cdot(3\cdot7^2)t^2+{(3\cdot7)^2\big)}^2=0\Rightarrow\\
  \Rightarrow &(t^2-49){(t^2+3\cdot7^2)}^2=0
\end{align*}
Now, it's time to recall that 
\[
  t=2x+7\Rightarrow x=\frac{t-7}2
\]
And use the solutions for $t^2$
Solution (1):
\[
  t=-7\Rightarrow x=\frac{-7-7}2=-7
\]
Solution (2):
\[
  t=7\Rightarrow x=\frac{7-7}2=0
\]
The rest:
\[
  t^2=-3\cdot7^2\Rightarrow t=\pm7i\sqrt3\Rightarrow x=-\frac{7\pm7i\sqrt3}2
\]
\end{document}
