Basic number theory curiosity

Question

Let $a,b $ be positive integers, and consider the set $D := \{am + bn: m,n \in \mathbb{Z} $ and $am + bn > 0 \}$. Let $d$ be the minimum integer in $D$. Is it true that $D = \{kd : k \in \mathbb{Z^{+}}\}$?

Answer 1

Yes. If $t\in D$, we can write $t=kd+r$ where $0\le r<d$. Note that, writing $r=t-kd$, we see that $r$ also has the form $am+nb$ for some $m,n\in\mathbb Z$. Hence, if $r\ne 0$, then $r\in D$, contradicting the minimality of $d$. This shows $D$ is a subset of $\{kd\mid k\in\mathbb Z^+\}$. On the other hand, if $d=am+bn$, then $kd=a(km)+b(kn)$, so $kd\in D$ as well, showing the other inclusion.

As mentioned in the comments, $d$ is the greatest common divisor of $a,b$, and all one ought to further verify to prove its existence is that $D\ne\emptyset$. But this is easy.