$f$ is continuous on $[0,1]$, $f(0)=f(1)$,$n\in \mathbb N$ constant $\Rightarrow$ There exists $x\in[0,1-\frac{1}{n}]$ : $f(x+\frac{1}{n})=f(x).$ I've been trying to prove this statement:

Let $f$ be a continuous function on $[0,1]$ such that $f(0)=f(1):=c$, and let $n$ be a natural constant. Prove that there exists $x\in[0,1-\frac{1}{n}], x\in \mathbb R$ such that $f(x+\frac{1}{n})=f(x).$

I tried to use the Intermediate Value Theorem by defining $g(x)=f(x+\frac{1}{n})-f(x)$. I could see that $g(0)=f(\frac{1}{n})-c$ and $g(1-\frac{1}{n})=c-f(1-\frac{1}{n})$ yet I could not figure out how to prove that $\mathrm s\mathrm i\mathrm g\mathrm n (g(0)\cdot g(1-\frac{1}{n}))=-1.$
Edit: Please notice that this statement is not (necessarily) true for every n. The statement is about some constant n.
Thank you and have a good day!.
 A: Let
$$g(x)=f\left(x+\frac{1}{n}\right)-f(x)$$
Then we need to prove that $\exists x$ so that $g(x)=0$.
$$g(0)=f\left(\frac{1}{n}\right)-c$$
$$g\left(\frac{1}{n}\right)=f\left(\frac{2}{n}\right)-f\left(\frac{1}{n}\right)$$
$$g\left(\frac{2}{n}\right)=f\left(\frac{3}{n}\right)-f\left(\frac{2}{n}\right)$$
$$\dots$$
$$g\left(\frac{n-2}{n}\right)=f\left(\frac{n-1}{n}\right)-f\left(\frac{n-2}{n}\right)$$
$$g\left(\frac{n-1}{n}\right)=c-f\left(\frac{n-1}{n}\right)$$
If one of the above values are $0$, then we are done. So suppose that all of them are different from $0$. But then we can sum them:
$$g\left(\frac{0}{n}\right)+g\left(\frac{1}{n}\right)+\dots+g\left(\frac{n-1}{n}\right)=0$$
Because none of the values on the LHS are zero, then there must be at least $1$ negative and at least $1$ positive. But this means that we are done: if $g(a)<0$ and $g(b)>0$ then $\exists d$ between $a$ and $b$ with $g(d)=0$, because $g$ is continuous.  
Remark1: And this statement is true for all $n \in \mathbb{N}_+$
Remark2: I had the same exercise 1 or 2 months ago. I wanted to make a question about it, but I figured out the solution while I was writing the question :)  
An extra exercise: Construct a function $f \in C[0,1]$, $f(0)=f(1)$ which does not have a chord with length $2/5$: So $\nexists x \in [0, 1-2/5]$ so that $f(x)=f(x+2/5)$. This was an additional excercise in my worksheet, but I could not solve it.
