Let $k$ divides both $m+2n$ and $3m+4n$, prove $k$ also divides $m$ and $2n$

Question

For any natural numbers $m$, $n$ and $k$, such that $k$ divides both $m + 2n$ and $3m+4n$, $k$ must be a common divisor of $m$ and $2n$.

I can confirm this postulation by putting trial and error values, say $m=10$, $n=5$, $m + 2n = 20$ & $3m + 4n = 50$.

$k$ can be $10$ which divides $m$ & $2n$.

But, is there a way to prove it more elegantly without using trial and error method?

Answer 1

If k|m+2n and k|3m+4n, then k|(3m+4n)-(m+2n), which means k|2m+2n. Subtracting k|m+2n from this result gives k|m. Subtracting this from k|m+2n results in k|2n. QED.