I'm having trouble solving this:
Let continuous function $f~:~[0,~1] \rightarrow \mathbb{R}$ have equal values at 0 and 1 ($f(0) = f(1)$). Prove that for any $\alpha$ of the form $1/n, n \in \mathbb{N}$ the equation (relative to $x$) $f(x+\alpha) = f(x)$ has a solution.
In addition, if $\alpha$ is not represented in the form $1/n$, show that we can present a function of such kind that would not have solutions for the equation above.
I believe we initially need to show that $g(x)= f(x+\alpha)-f(x)$ cannot be of the same sign for the whole close interval $[0, 1-\alpha]$, somehow making use of the Intermediate value theorem, but I have problems deriving that.
