1. ## Divisibility

Question: If $a\mid{b}$ and $c\mid{d}$, must $(a+c)\mid(b+d)?$

Counterexample: $2\mid{4}$ and $3\mid{9}$, but $$5\nmid{13}$$.

However, I want to argue along the lines of if $a\mid{b}$ and $c\mid{d}$, then $b = aq$ and $d = cq'$
for some integers [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] . How should we proceed from there?

2. I don't understand your question. The counterexample is proof enough that the theorem is false. Are you seeking a way of finding solutions?

3. Originally Posted by Media_Man
I don't understand your question. The counterexample is proof enough that the theorem is false. Are you seeking a way of finding solutions?
Is the counterexample the only way to show that the statement is false? I was hoping that if we proceed as if we were to prove it,
we could derive a contradiction. Perhaps I'm expecting the wrong thing, but I do of course realise that a counterexample is enough
to be prove it false. Also, now that you mention it, how can we find the set of values for which it holds?

4. ## Finding Solutions

how can we find the set of values for which it holds?
Problem: Choose arbitrary a,c and find values b,d fitting the following three criteria: a|b, c|d, and a+c|b+d

Solution: Restate the three criteria as b=aq, d=cq', b+d=(a+c)n (as you have correctly done in your first post). Substitute (1) and (2) into (3) to get aq+cq'=(a+c)n. This can be rearranged as a/c=(n-q')/(q-n). The left and right are both equal rational numbers, therefore the numerators and denominators are scalar multiples: as=n-q', cs=q-n. So q=n+cs and q'=n-as. Substituting back into (1) and (2), b=an+sca and d=cn-sca. So given 4 independent parameters a,c,n,s, b and d can be gotten by this formula to satisfy your three criteria. Proving this captures ALL solutions is probably trickier, but someone else might like to tackle it.