, 2011

Choose any positive rational $p, q$ and let $x = \frac{p+q}{2}$ and $y = \frac{p-q}{2}$. Then
$$f(p) - f(q) = 4\sqrt{f(x)f(y)} > 0,$$
so $f(p) > f(q)$ and hence $f$ is strictly increasing.

Now we show that $f$ has values arbitrarily close to $0$, i.e., for any $\epsilon \in \mathbb{Q}^+$, we can find $x$ such that $f(x) < \epsilon$. Substituting $x = 2y$ gives
$$f(3y) - f(y) = 4\sqrt{f(2y)f(y)} > 4\sqrt{f(y)f(y)} = 4f(y),$$
and hence $f(3y) > 5f(y)$ or $f(y) < \frac{f(3y)}{5}$. Substituting this into itself gives $f(y) < \frac{f(3^2y)}{5^2}$, and repeating gives $f(y) < \frac{f(3^ky)}{5^k}$. Now let $y = \frac{1}{3^k}$, giving $f(\frac{1}{3^k}) < \frac{f(1)}{5^k}$. Since $f(1)$ does not depend on $k$, we can choose $k$ sufficiently large so that $\frac{f(1)}{5^k}$ is less than $\epsilon$ and hence $y$ satisfies the condition.

Consider the original equation again, and consider what happens as $f(y)$ tends to zero. $f(x)$ stays fixed so $4\sqrt{f(x)f(y)}$ can be made arbitrarily small. Also, $f(x)$ lies between $f(x-y)$ and $f(x+y)$, so for any $\delta$ we can choose a $y$ such that $f(x) - \delta < f(x-y) < f(x) < f(x+y) < f(x) + \delta$.

Now let $x = z + \epsilon$ and $x = z - \epsilon$. The original equation becomes
$$f(2z) = 4\sqrt{f(z + \epsilon)f(z - \epsilon)} + f(2\epsilon).$$
By choosing $\epsilon$ sufficiently small, we can ensure that $f(\epsilon)$ is arbitrarily close to zero and $f(z + \epsilon)$ and $f(z - \epsilon)$ are arbitrarily close to $f(z)$. The right hand side can thus be made arbitrarily close to $4f(z)$, and it follows that the left hand side must be $4f(z)$.

We can extend $f(2z) = 4f(z)$ to $f(nz) = n^2 f(z)$ for all positive integers $n$. For $n = 1$ and $n = 2$ we already have it. Suppose it holds up to some $n$. We now show that it also holds for $n + 1$. Let $x = nx, y = x$, giving
$$\begin{aligned} f((n + 1)x) &= 4\sqrt{f(nx)f(x)} + f((n - 1)x) \\ &= 4\sqrt{n^2 f(x)f(x)} + (n - 1)^2 f(x) \\ &= 4n f(x) + (n - 1)^2 f(x) \\ &= (n + 1)^2 f(x). \end{aligned}$$
The result follows by induction.

We can further extend this to positive rational coefficients. Let $n, m$ be positive integers. $n^2 f(x) = f(m \cdot \frac{n}{m} \cdot x) = m^2 f(\frac{n}{m} \cdot x)$, from which it follows that $f(\frac{n}{m} \cdot x) = (\frac{n}{m})^2 f(x)$.

Now let $f(1) = a$. By the above, $f(x) = a x^2$ for all positive rationals $x$. Conversely, it is trivial to substitute this into the original equation to confirm that it is a valid solution, for any $a \in \mathbb{Q}^+$.