\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Longlist 1985/19}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 124}
\maketitle

\section*{Problem}
Solve over $\RR$ the system of simultaneous equations
\begin{align*}
  \sqrt x - \frac 1y - 2w + 3z &= 1, \\
  x + \frac{1}{y^2} - 4w^2 - 9z^2 &= 3, \\
  x \sqrt x - \frac{1}{y^3} - 8w^3 + 27z^3 &= -5, \\
  x^2 + \frac{1}{y^4} - 16w^4 - 81z^4 &= 15.
\end{align*}

\section*{Video}
\href{https://www.youtube.com/watch?v=V_tTeGYaVAk&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/V\_tTeGYaVAk}}

\section*{External Link}
\url{https://aops.com/community/p2014940}

\newpage

\section*{Solution}
Let $a = \sqrt x$, $b = -1/y$, $c = 2w$, $d = -3z$.
\begin{align*}
a + b - c - d &= 1 \\
a^2 + b^2 - c^2 - d^2 &= 3 \\
a^3 + b^3 - c^3 - d^3 &= -5 \\
a^4 + b^4 - c^4 - d^4 &= 15.
\end{align*}
This condition is the same as saying
\[ a^n + b^n + (-1)^n + (-1)^n = c^n + d^n + (-2)^n + 1^n
\qquad n = 1, 2, 3, 4. \]
which is equivalent to saying the multiset
$\{a,b,-1,-1\}$ is the same as the multiset $\{c,d,-2,1\}$,
(because Newton's formulas imply
the polynomials with these roots have the same coefficients).
Therefore, $\{a,b\} = \{-2,1\}$ while $c=d=-1$.

Going back, with $a > 0$ this gives only one solution, which evidently works:
\[ (x,y,w,z) = (1, 1/2, -1/2, 1/3). \]

\end{document}
