For function $f:\mathbb Z^+ \to \mathbb R$ and coprime positive integers $p,q$ ; define $f_p,f_q$ as $$f_p(x)=f(px)-f(x), f_q(x)=f(qx)-f(x) \space \space (x\in\mathbb Z^+)$$$f$ satisfies following conditions. $ $ $ $ $(i)$ $ $ for all $r$ that isn't multiple of $pq$, $f(r)=0$ $ $ $ $ $(ii)$ $ $ $\exists m\in \mathbb Z^+$ $ $ $s.t.$ $ $ $\forall x\in \mathbb Z^+, f_p(x+m)=f_p(x)$ and $f_q(x+m)=f_q(x)$ Prove that if $x\equiv y$ $ $ $(mod m)$, then $f(x)=f(y)$ $ $ ($x, y\in \mathbb Z^+$).