For any integers $a, b, d$, if $d | a$ and $d | b$, then $d$ | $(3a + 5b)$.

Question

I guessed the hypothesis and the conclusion and I just have to prove it.

Hypothesis: $d$ | $a$ and $d$ | $b$
Conclusion: $d$ | $(3a + 5b)$

If $d$ | $a$, then $a = dk$ for some integer $k$.
If $d$ | $b$, then $b = dl$ for some integer $l$.
then $3a+5b$ = $d ⋅ (3k+5l)$

Is there more to this proof or is this all I need to write. Need help please. Thank you in advance.

Answer 1

You:

"If d | a, then a=dk for some integer k.

If d | b, then b=dl for some integer l.

then 3a+5b = d⋅(3k+5l)"

Me:

$3k+5l$ is some integer.

Since $3a+5b = d*(3k+5l)$ for some integer, $3k + 5l$, $d|3a + 5b$.

......

More general, more useful, and more powerful: if $d|a$ and $d|b$ then $d$ divides any $\pm ak$ and any $\pm bm$ and any $\pm ak \pm bm$ for ALL possible combinations where $k,m$ are integers (positive, negative, or zero). Can you prove that.