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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 $x$.
}

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\textbf{Solution}

Let $k$ be the least upper bound for $|f(x)|$. %, so $k > 0$.
Suppose $|g(y)| > 1$.
Then
$$2k \ge |f(x + y)| + |f(x - y)| \ge |f(x + y) + f(x - y)| = 2 |g(y)||f(x)|,$$
so $|f(x)| \le k/|g(y)|$.
In other words, $k/|g(y)|$ is an upper bound for $|f(x)|$ which is less than $k$.
Contradiction.

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