Problem: Let $a_0$, $b_0$, $c_0$ be complex numbers, and define
\begin{align*}
  a_{n+1} &= a_n^2 + 2b_nc_n \\
  b_{n+1} &= b_n^2 + 2c_na_n \\
  c_{n+1} &= c_n^2 + 2a_nb_n
\end{align*}
for all nonnegative integers $n$.
Suppose that $\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022$ for all $n \ge 0$.
Prove that \[ |a_0|^2 + |b_0|^2 + |c_0|^2 \leq 1. \]

\end{enumerate}

Solution: For brevity, set $s_n \coloneq |a_n|^2 + |b_n|^2 + |c_n|^2$.
Note that the $s_n$ are real numbers.
\begin{claim*}
  [Key miraculous identity]
  We have
  \[ s_{n+1} - s_n^2 = 2 |a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n}|^2. \]
\end{claim*}
\begin{proof}
  We prove this by mechanical calculation. First,
  \begin{align*}
    s_{n+1} &= \left\lvert a_n^2+2b_nc_n \right\rvert^2 +
    \left\lvert b_n^2+2c_na_n \right\rvert^2 +
    \left\lvert c_n^2+2a_nb_n \right\rvert^2 \\
    &= \sum_{\text{cyc}} \left\lvert a_n^2+2b_nc_n \right\rvert^2 \\
    &= \sum_{\text{cyc}} (a_n^2+2b_nc_n)(\ol{a_n}^2 + 2\ol{b_n}\ol{c_n}) \\
    &= \sum_{\text{cyc}} \left( |a_n|^4 + 2 \ol{a_n}^2 b_nc_n
      + 2a_n^2 \ol{b_n} \ol{c_n}  + 4 |b_n|^2 |c_n|^2 \right) \\
      &= s_n^2 + 2 \sum_{\text{cyc}}
      \left( \ol{a_n}^2 b_nc_n
      + a_n^2 \ol{b_n} \ol{c_n} + |b_n|^2 |c_n|^2 \right).
  \end{align*}
  Meanwhile,
	\begin{align*}
		|a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n}|^2
		&= \left( a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n} \right)
		\left( \ol{a_n} b_n + \ol{b_n} c_n + \ol{c_n} a_n \right) \\
		&= |a_n|^2 |b_n^2| + a_n \ol{b_n}^2 c_n + a_n^2 \ol{b_n} \ol{c_n} \\
		&\qquad + \ol{a_n} b_n^2 \ol{c_n} + |b_n|^2 |c_n|^2 + a_n b_n \ol{c_n}^2 \\
		&\qquad + \ol{a_n}^2 b_n c_n + \ol{a_n} \ol{b_n} c_n^2 + |a_n|^2 |c_n|^2
	\end{align*}
	which exactly matches the earlier sum, term for term.
\end{proof}

In particular, $s_{n+1} \ge s_n^2$, so applying repeatedly,
\[ s_n \ge s_0^{2^n}. \]
Hence if $s_0 > 1$, it follows $s_n$ is unbounded, contradicting
$\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022$.

\begin{remark*}
The originally intended solution was to capture all three
recursions in the following way.
First, change the recursion to
\begin{align*}
  a_{n+1} &= a_n^2 + 2b_nc_n \\
  c_{n+1} &= b_n^2 + 2c_na_n \\
  b_{n+1} &= c_n^2 + 2a_nb_n
\end{align*}
which is OK because we are just rearranging the terms in each triple.
Then if $\omega$ is any complex number with $\omega^3 = 1$, and we define
\[ z_n \coloneq a_n + b_n \omega + c_n \omega^2, \]
the recursion amounts to saying that $z_{n+1} = z_n^2$.
This allows us to analyze $|z_n|$ in a similar way as above,
as now $|z_n| = |z_0|^{2^n}$.
\end{remark*}