TITLE: How to find period of $f$ if $f(x+13) + f(x+630) = 0$
QUESTION [7 upvotes]: Let $f:\Bbb{R}\to\Bbb{R}$ be a periodic function with period $T$. The question was to find the (fundamental) period given the following relation.
$$ f(x+13) + f(x+630) = 0 $$
Now, the given method is:
$$\begin{align} 
&f(x+13) + f(x+630) = 0 \\
\implies &f((x+617)+13) + f((x+617)+630) = 0 \\
\implies &f(x+630) + f(x+1247) = 0 \\
\end{align}$$
Subtracting this equation from the 'original', we have:
$$ f(x+1247) = f(x+13) $$
So, the period is $ 1234 $. 
All this is fine. However, my doubt is this: How do we know that $1234$ is the fundamental and not just any period?

REPLY [4 votes]: The prime factors of $1234$ are $2$ and $617$. Since $f(x+617)=-f(x)$, $617$ isn't a period, and we need only rule out $2$. In fact it appears to me that $2$ could be the fundamental period of $f$: just take $f(x)=\sin\pi x$, for instance.. (And that's assuming that the period is an integer, which of course it need not be.)

REPLY [2 votes]: We don't. We could as well have $1234=nT$ with some $n\in\mathbb N$. However, as long as $f$ is not the zero function, we know that $617$ is not a multiple of $T$, but that only gives us that $n$ must be odd.