\documentclass[11pt]{scrartcl}
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\begin{document}
\title{H3087478}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 139}
\maketitle

\section*{Problem}
Positive integers $a$, $b$, $c$, $d$ satisfy the conditions
\begin{align*}
  a+b+c+d &= (a-b)^2 \\
  ab &= cd.
\end{align*}
Show that $4c+1$ is a square.

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

\section*{External Link}
\url{https://aops.com/community/p27901399}

\newpage

\section*{Solution}
The product condition implies that we can choose integers $u,v,x,y > 0$ with
\[ a = ux, \qquad b = vy, \qquad c = vx, \qquad d = uy. \]
In that case we can write
\begin{align*}
  a+b+c+d &= (u+v)(x+y) \\
  (a-b)^2 &= (ux-vy)^2.
\end{align*}
Define
\begin{align*}
  \alpha &\coloneqq u+v \\
  \beta &\coloneqq x+y.
\end{align*}
Then the given equation becomes
\[ (x\alpha-v\beta)^2 = \alpha \beta. \]
This lets us set $\alpha = \gcd(\alpha,\beta) m^2$,
$\beta = \gcd(\alpha,\beta) n^2$ for coprime $m,n > 0$.
This implies that
\[ xm^2 - vn^2 = \pm mn
  \implies x \left( \frac mn \right)^2 \mp \frac mn - v = 0. \]
So in particular, the quadratic equation
\[ x \theta^2 \mp \theta - v = 0 \]
in the variable $\theta$ has a rational solution $\frac mn$.
Therefore the discriminant
\[ (\mp1)^2 - 4vx = 1 + 4vx  = 4c + 1\]
is a perfect square, as desired.

\begin{remark*}
  For any positive integer $k$, an example of a satisfying quadruple is
  \[ (a,b,c,d) = \left( k^2, (k+1)^2, k(k+1), k(k+1) \right) \]
  which corresponds to
  \begin{align*}
    (u,v,x,y) &= (k,k+1,k,k+1) \\
    \alpha = \beta &= 2k + 1 \\
    m = n &= 1.
  \end{align*}
\end{remark*}

\begin{remark*}
  The problem is actually false if the hypothesis $a,b,c,d > 0$ is dropped,
  but only in the situation
  where the variables $\alpha$ and $\beta$ could equal zero.
  For example, this happens if $(a,b,c,d) = (1,1,-1,-1)$.
\end{remark*}

\end{document}
