Show that $S_{x,y} ⊆ \{n : n ∈ \mathbb Z\textrm{ and }d|n\}$.

Question

For $x, y ∈ \mathbb Z$ we define the set: $S_{x,y} = \{mx + ny : m, n ∈ \mathbb Z\}$. Let $d = \gcd(x, y)$.

So I know that

$x = k_1\cdot d$

$y = k_2\cdot d$ (for some $k_1, k_2 ∈ \mathbb Z$)

So $mx+my = m(k_1\cdot d) + n(k_2\cdot d) = d(m\cdot k_1 + n\cdot k_2)$

Thus, $d|mx+ny$.

But does that imply $d|n$?

Answer 1

I can't comment (not enough reputation). You did a good job, and you indeed proved that $S_{x,y}\subset \{n\in\mathbb N : d\mid n\}$.