Putnam 1947 
Topic: Analysis Algebra

Problem: A real valued continuous function satisfies for all real $x$ and $y$ the functional equation 
\[ f(\sqrt{x^2 + y^2}) = f(x)f(y). \]
Find the value of f(x) in terms of f(1). \]

Solution: A slight qualification in the statement of the problem is needed since the real valued continuous function $f(x) \equiv 0$ satisfies the functional equation for all real $x$ and $y$, but does not satisfy the relation $f(0) = [f(1)]^0$, since $0^0$ is undefined.

Assume then that for some $y_0$, $f(y_0) \neq 0$. Since
\[ f(x)f(y_0) = f(\sqrt{x^2 + y_0^2}) = f(-x)f(y_0), \]
we have $f(x) = f(-x) = f(|x|)$ for all $x$. We now show by induction that for any positive integer $n$ and any real number $x$, we have
\[ f(\sqrt{n}x) = [f(x)]^n. \]
This is certainly true for $n = 1$, and assuming it true for $n = k$ we have
\[ f(\sqrt{k+1}x) = f(\sqrt{k+1}|x|) = f(\sqrt{(\sqrt{k}x)^2 + x^2}) = f(\sqrt{k}x)f(x) \]
\[ = [f(x)]^k f(x) = [f(x)]^{k+1}. \]
Therefore (1) is true for all positive integers $n$.

If $p$ and $q$ are non-zero integers, then
\[ f(p) = f(|p|) = f(\sqrt{p^2 \cdot 1}) = [f(1)]^{p^2} \]
and
\[ f(|p|) = f\left(\sqrt{\frac{q^2}{q^2}|p|^2}\right) = \left[f\left(\frac{|p|}{q}\right)\right]^{q^2}. \]
From these two relations it follows that
\[ \left[f\left(\frac{p}{q}\right)\right]^{q^2} = [f(1)]^{p^2}. \]
If $f(1) > 0$ it then follows that
\[ \left[f\left(\frac{p}{q}\right)\right] = [f(1)]^{p^2/q^2}; \]
that is, the required equation is valid for all rational values of $x$ except, perhaps, $x = 0$. By continuity it follows for all values of $x$.

If $f(1) = 0$, then (2) implies that $f(\frac{p}{q}) = 0$ for all non-zero integers $p$ and $q$, and thus $f(x) = 0$ for all rational $x$, hence for all real $x$.

Finally we show that $f(1) < 0$ is impossible. If $p$ is even and $q$ is odd, equation (2) implies that $f(\frac{p}{q}) > 0$. Hence $f(x) > 0$ for a dense set of $x$, and therefore $f(x) \geq 0$ for all $x$; in particular, $f(1) \geq 0$. This makes the final answer \boxed{f(1)^{x^2}}.