Thread: Divisibility

Question: If and , must

Counterexample: and , but [tex]5\nmid{13}[/Math].

However, I want to argue along the lines of if and , then and
for some integers [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] . How should we proceed from there?

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

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?

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.