**Problem**: Given two positive reals $ \alpha $ and $ \beta $, show that there is a continuous function $ f $ that satisfies $ f(x) = f(x+\alpha)+f(x+\beta) $.

**Solution**: There are several special cases that are interesting to look at before we make a guess as to what type of function $ f $ will be. First we consider the case $ \alpha = \beta = 1 $. This immediately gives

$ f(x) = 2f(x+1) $.

It should not be too difficult to guess that $ f(x) = 2^{-x} $ is a solution to this, as well as any constant multiple of it. Now try $ \alpha = -1 $ and $ \beta = -2 $, resulting in

$ f(x) = f(x-1)+f(x-2) $.

Looks a lot like Fibonacci, right? In fact, one possible function is just $ f(x) = \phi^x $. That's pretty convenient.

Notice how both of these illuminating examples are exponential functions, which leads us to guess that our function will be exponential as well. So, following this track, we set

$ f(x) = a^x $

so we simply need to solve

$ a^x = a^{x+\alpha}+a^{x+\beta} $

$ a^x = a^x\left(a^{\alpha}+a^{\beta}\right) $.

Unfortunately, the equation $ a^{\alpha}+a^{\beta} = 1 $ does not always have a solution (take $ \alpha = 1 $ and $ \beta = -1 $ for example). But that's ok and I'll worry about it some other time. In any case we have found a function for whenever the equation $ a^{\alpha}+a^{\beta} = 1 $ has a solution in the reals.