number theory - Arithmetic progression divisibility

Suppose $b|a$ and $\frac{a}{b} \neq \frac{v}{y}$, $a, b, v, y \in \mathbb{N}$ arbitrary. Is there a nice clean intuitive proof to show that it is never true that $b+yk|a+vk$ for all $k \in \mathbb{N}$? Or is it true sometimes after all (I strongly feel like not)?

Answer

Assume for contradiction that for all $k\in \Bbb N$, $a+vk|b+yk$ and consider the sequence $u_k\in \Bbb Z$ such that $u_k(a+vk) = b+yk$. Then since $u_k = {b+yk\over a+vk} \to {y\over v}$ as $k$ goes to $+\infty$, $u_k$ approaches ${y\over v}$ arbitrary close, which therefore must be an integer, and thus $v|y$. Let $u \in \mathbb{Z}$ be such that $y = vu$. Then for all $k\in \Bbb N$ we have $a+vk |b+vuk$ and therefore also $a+vk |b-au$. However since $a+vk$ is unbounded and $b-au$ does not depend on $k$, what follows is that $b-au = 0$, which rewrites as ${a\over b} = {v \over y}$.

The contrapositive of this is your statement.