Values of $a$ s.t. for all continuous $f$ with $f(0)=f(1)$ there exists $x$ s.t. $f(x+a) = f(x)$ 
Determine all $a\in[0,1]$ such that for ${\it every}$ continuous function $f:[0,1]\to \mathbb{R}$ with $f(0)=f(1)$ there exists at least one $x$ where $f(x) = f(x+a)$.

First of all, $a=0,1/2,1$ are obviously good values. Secondly, no $1/2<a<1$ works (look at $f = \sin(2\pi x)$).
I remember trying to solve this several years ago. I think I started by dividing $[0,1]$ into regions where $f\geq 0$ and $f<0$. Then studied each interval and all pairs of intervals with the same sign. However, this seems like a problem that should have a simple solution and I would love to see it.
 A: The set you're looking for is
$$ \{0\} \cup \left\{\frac{1}{n}\right\}_{n\in \mathbb N}. $$
My proof (not a really "simple" one!) is going to be somewhat "visual"; I'm also going to only consider (WLOG) functions such that $ f(0) = f(1) = 0 $.
If $ a = 1/n $ for a natural number $ n $, you can partition the interval $ [0,1] $ into $ n $ equally broad "bands". Let $ f $ be any continuous function; one can show that if the function is not "overall strictly increasing" ("o.s.i."; I made up this terminology) over each band, in the sense that $ f(ka) < f((k+1)a) \; \forall k $, then $$ (*)\qquad f(x) = f(x+a) $$ for some $ x $. To see this, draw the graph of $ f(x+a) $, by moving the graph of $ f(x) $ forward by $ a $, and notice that the condition $ (*) $ means that the graph of $ f(x) $ should cross that of $ f(x+a) $, which will necessarily happen if the "o.s.i." condition isn't fulfilled (you can convince yourself this is true by either visually picturing the described situation or applying the intermediate value theorem to $ f(x+a) - f(x) $). But clearly, a function satisfying this o.s.i. condition cannot simultaneously satisfy $ f(0) = f(na) = f(1) $. Therefore $ a $ belongs to your set.
On the other hand, if $ a \ne 1/n $ for a natural $ n $, you should be able to explicitly conjure up a function $ f $ for which $ f(x) \ne f(x + a) $ for any $ x \in [0,1] $. I came up with the following: let $ 1 = na + \delta $, with $ 0 < \delta < n $ by the hypothesis on $ a $. I'm going to define the value of $ f $ at certain points, and then $ f $ may be build by simply joining them with line segments.
I'm choosing an arbitrary $ \varepsilon > 0 $ and then setting
$$ \begin{cases}f(ka + \delta) = (n-k)\varepsilon \quad&\mathrm{for}\; k = 0,1,\ldots,n \\ f(ka) = -k\varepsilon \quad&\mathrm{for}\; k = 0,1,\ldots,n.\end{cases} $$ This function should not cross with its own shifted version. The gist is that in this case the above "band partition" isn't able to account for the whole $ [0,1] $ interval, so you can avoid crossing the shifted-function's graph by following a suitably oscillating path.
A: Decomposition into Fourier series
\begin{eqnarray*}
f(x) &=&\sum_{k}f_{k}\exp [2\pi ikx],\;x\in \lbrack 0,1] \\
f(x+a) &=&\sum_{k}f_{k}\exp [2\pi ik(x+a)]
\end{eqnarray*}
We must have
\begin{equation*}
\exp [2\pi ika]=1
\end{equation*}
for some $k.$ Fix $k$. Then this is the case for $a=1/n$ for some suitable
integer $n$. Then
\begin{equation*}
f_{k}(x)=f_{k}\exp [2\pi ikx]
\end{equation*}
fills the bill. For real $f(x)$ take the real part of $f(x)$
\begin{equation*}
f(x)=\sum_{k}\{f_{k}\exp [2\pi ikx]+\bar{f}_{k}\exp [-2\pi ikx]\}
\end{equation*}
