Problem 1631

Solution:
Consider $f(0) = f(0+0) = \frac{1}{f(0)} + \frac{1}{f(0)}$ which gives $[f(0)]^2 = 2$. Thus, $f(0) = \pm \sqrt{2}$.

Let $x = f(2011)$.

If $f(0) = \sqrt{2}$ then $x = f(2011) = f(2011+0) = \frac{1}{f(2011)} + \frac{1}{f(0)} = \frac{1}{x} + \frac{1}{\sqrt{2}}$. So, $x = \frac{\sqrt{2} + x}{\sqrt{2} x}$ which yields $\sqrt{2} x^2 - x - \sqrt{2} = 0$.

Solving for $x$ using the quadratic formula yields:
$$x = \frac{1 \pm \sqrt{1 - 4(\sqrt{2})(-\sqrt{2})}}{2 \sqrt{2}} = \frac{1 \pm 3}{2 \sqrt{2}}$$
Hence, if $f(0) = \sqrt{2}$ then either $f(2011) = \sqrt{2}$ or $f(2011) = -\frac{\sqrt{2}}{2}$.

However, suppose $f(2011) = -\frac{\sqrt{2}}{2}$. Consider $f(0) = f(2011 + (-2011)) = \frac{1}{f(2011)} + \frac{1}{f(-2011)}$ which implies that $\sqrt{2} = -\sqrt{2} + \frac{1}{f(-2011)}$. Thus, $f(-2011) = \frac{\sqrt{2}}{4}$.

But if we consider $f(-2011) = f(-2011+0) = \frac{1}{f(-2011)} + \frac{1}{f(0)}$, this means that $\frac{\sqrt{2}}{4} = 2\sqrt{2} + \frac{1}{f(0)}$.
Thus, $f(0) = -\frac{2\sqrt{2}}{7}$ which is a contradiction. Thus for $f(0) = \sqrt{2}$, $f(2011) = \sqrt{2}$.

If $f(0) = -\sqrt{2}$ then $x = f(2011) = f(2011+0) = \frac{1}{f(2011)} + \frac{1}{f(0)} = \frac{1}{x} - \frac{1}{\sqrt{2}}$. So, $x = \frac{\sqrt{2} - x}{\sqrt{2} x}$ which yields $\sqrt{2} x^2 + x - \sqrt{2} = 0$.

Solving for $x$ using the quadratic formula yields:
$$x = \frac{-1 \pm \sqrt{1 - 4(\sqrt{2})(-\sqrt{2})}}{2 \sqrt{2}} = \frac{-1 \pm 3}{2 \sqrt{2}}$$
Hence, if $f(0) = -\sqrt{2}$ then either $f(2011) = -\sqrt{2}$ or $f(2011) = \frac{\sqrt{2}}{2}$.

However, suppose $f(2011) = \frac{\sqrt{2}}{2}$. Consider $f(0) = f(2011 + (-2011)) = \frac{1}{f(2011)} + \frac{1}{f(-2011)}$ which implies that $-\sqrt{2} = \sqrt{2} + \frac{1}{f(-2011)}$. Thus, $f(-2011) = -\frac{\sqrt{2}}{4}$.

But if we consider $f(-2011) = f(-2011+0) = \frac{1}{f(-2011)} + \frac{1}{f(0)}$, this means that $-\frac{\sqrt{2}}{4} = -2\sqrt{2} + \frac{1}{f(0)}$.
Thus, $f(0) = \frac{2\sqrt{2}}{7}$ which is a contradiction. Thus for $f(0) = -\sqrt{2}$, $f(2011) = -\sqrt{2}$.

So the possible values for $f(2011)$ are $\pm \sqrt{2}$.