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\textbf{Problem} [B2 from {\small IMO} 1972]

\emph{%
$f$ and $g$ are real-valued functions defined on the real line.
For all $x$ and $y$, $$f(x + y) + f(x - y) = 2 f(x) g(y).$$
$f$ is not identically zero and $|f(x)| \le 1$ for all $x$.
Prove that $|g(x)| \le 1$ for all $x$.
}

\bigskip
\textbf{Tom Hales' Solution}

Note first that $|f(x)| |g(y)|^l \le 1$ for all $l \ge 0$, by
induction on $l$.
For the induction step:
\begin{eqnarray*} 
2 |f(x)| |g(y)|^{l+1} & = & |2 f(x) g(y)| |g(y)|^l \\
  & = & |f(x + y) + f(x - y)| |g(y)|^l \\
  & \le & |f(x + y)| |g(y)|^l + |f(x - y)| |g(y)|^l \\
  & \le & 2
\end{eqnarray*}
Now suppose that $|g(y)| > 1$ for some $y$. We know $f(z) \not= 0$ for some $z$, but
then $|f(z)| |g(y)|^l \to \infty$ as $l \to \infty$, contradicting the bound.

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