Proof that any integer of the fundamental period, is also a period of the function? How can it be proved that for a given $f(x)$ with a period $p$, that for any integer $n=1,2,\dots$, the product $np$ is also an period of the function $f(x)$?
I know that the definition of a periodic function is that there is some non-zero value, $p$ (called the period) such that $f(x+p) = f(x)$ for all $x$.
So I'm guessing that the proof may starts off with something like 
$f(nx) = f(nx +np) \dots $ ?
But I'm really no sure how to prove this one.
 A: By the definition of periodic, there is a number $p$ so that for every $x$,
$$ f(x+p)=f(x). $$
We can then proceed essentially by induction: we have that $1p$ is a period, by definition. Now suppose that $np$ is a period, so $ f(x+np)=f(x) $. But then
$$ f(x+(n+1)p) = f((x+np)+p) = f(x+np) = f(x), $$
so if $np$ is a period, so is $(n+1)p$. Thus induction gives that $np$ is a period for every positive integer $n$. It is clear that $-p$ is a period if and only if $p$ is: if $p$ is a period,
$$ f(x) = f((x-p)+p) = f(x-p), $$
so $-p$ is, and vice versa.
A: If we want to be formal, we can do an induction on $n$. We do the induction step.
Let $p$ be a period of $f$, and suppose that we know that $kp$ is a period of $f$. We want to show that $(k+1)p$ is a period of $f$. Compute. For any $x$, $$f(x+(k+1)p)=f((x+kp) +p=f(x+kp)=f(x).$$
A: Prove $f(x+np)=f(x)$ by induction on $n$.
By definition, $f(x+1p)=f(x+p)=f(x)$.
So this is true for $n=1$.
Assume it true for $n$. Then
$$f(x(n+1)p)=f((x+np)+p)=f(x+np)=f(x)$$
So it is true for all $n$.
