Let $f$ be a continuous function from $\Bbb R$ to itself.
If $ f(m\pm n\pi)=0$ for all $m,n\in \Bbb Z$ show that $f=O$
I am unable to proceed in this case.
Sorry I couldn't do anything positive to show you all.
I tried with some examples but don't know what to do.
Hint: Here are the relevant facts:
If $\alpha \in \mathbb R$, then $\mathbb Z + \alpha\mathbb Z=\{m+n\alpha\ : m,n\in \mathbb Z\}$ is an additive subgroup of $\mathbb R$ .
An additive subgroup of $\mathbb R$ is either cyclic or dense.
$\mathbb Z + \alpha\mathbb Z$ is cyclic iff $\alpha$ is rational.
$\pi$ is irrational.