$[0,n]$ continuous function problem Let $n\in \mathbb{N}$. $f$ is a real valued function, defined and continuous on $[0,n]$ and $f(0)=f(n)$. Prove that there exists $x_1,x_2\in [0,n]$, so that $x_1-x_2=1$ and $f(x_2)=f(x_1)$.
Hints on how to approach this problem would be appreciated.
 A: Hint: Assuming you meant continuous functions from $[0,n]$ to $\mathbb{R}$, consider a new function
$$ g(x) = f(1+x)-f(x), \forall x\in[0,n-1]$$.
Try and prove that $g(x)$ is continuous and crosses zero. So, $\exists x\in[0,n-1]$ such that $f(1+x)=f(x)$. 
A: I think your question is incomplete; consider the map
$$f:\ [0,2]\ \longrightarrow\ \Bbb{R}^2:\ x\ \longmapsto (\cos(\pi x),\sin(\pi x)).$$
It is clear that $f$ is continuous and that $f(0)=f(2)$, but for $x_1,x_2\in[0,2]$ we have $f(x_1)=f(x_2)$ if and only if $x_1,x_2\in\{0,2\}$. In particular we cannot have $x_1-x_2=1$.
A: Just to expand on Shash's answer:
Notice that due to Rolle's theorem, we know there is $\xi \in (0,n)$ such that $f'(\xi)=0$. 
Suppose wlog that $f(\xi)$ is local maximum, and $\xi >1$.
In that case, $g(0)=f(1)-f(0)$ is larger than zero, because $1$ is on the way $\xi$ and the function is ascending.
That also means that $g(n-1)=f(n)-f(n-1)$ is negative due to a similar argument.
$g$ is continuous, $g(0)*g(n-1) <0$, this means (think bisection method) that there is $\psi \in (0,n-1)$ such that $g(\psi)=0$. this means that $f(1+\psi)=f(\psi)$, and we are done.
Disclaimer: this argument succeeds if $f(\xi)$ is local minimum / maximum.
In that case that it isn't (meaning $f''(x)=0$) then I'm not 100% sure that what you are trying to prove holds).
Disclaimer 2: The assumption that $\xi >1$ is important and non trivial. even if $f(\xi)$ is local extremum, if $\xi <1$ then my proof won't work.
