How would I formally prove that for the integers $a$, $b$, and $d$, if $d\mid da+b$, then $d\mid b$? Would a direct proof be the best option? If I do a direct proof I seem to get stuck pretty quickly... in fact I don't even know how to start...
Assume $d\mid da+b$ ... ?
This might look like a HW dump but I honestly just don't know how to continue...
Start out by assumption to proove the conditional. So now we know that $d | (da + b)$ and we want to proove $d | b$. Our assumption really means that there is an integer, call it $c$, such that $\frac{(da + b)}{d}$ is an integer. That means that $c=a + b/d$ is an integer. An integer plus something else must be an integer only if that something else is an integer. So because we know that $a$ is an integer and $b/d$ can be that "something else" we know that $b/d$ must be an integer. Well what does that mean? it means that $d|b$.
More precise formalization is left as an exercise.
Since $d\mid(ad+b)$, there exists an integer $n$ such that $nd = (ad+b)$. So $nd-ad = b$ and thus $d(n-a) = b$. But since $n$ and $a$ are integers, it follows that $d$ divides $b$.
Lemma: If $d|a$ and $d|b$ then $d|a+b$.
Proof: If $d|a$ and $d|b$ then there are integers $x_1$, $x_2$ such that $a=dx_1$and $b=dx_2$. Then $a+b=d(x_1+x_2)$.
Now to main statement. Assume $d|ad+b$. We know $d$ divides $-ad$ as $-ad=d(-a)$. Thus $d|(ad+b-ad)=b$, or $d|b$.