Here's the question you clicked on:
2bornot2b
If gcd(a,b)=1 then prove that gcd(ac,b)=gcd(c,b)
HINT: (a,b) = 1 means that there is some x and y such that ax + by = 1
perhaps start by writing \(d=gcd(ac,b)\) and \(e=gcd(b,c)\) and then show that \(d|e\) and \(e|d\)
How about this: http://math.stackexchange.com/questions/20889/
yes i guess that will do it. notice this fairly obvious fact actually takes a few lines to prove!
Yes, sat, it's one of the many things of number theory which are just intuitively plausible.
what do you mean @FoolForMath
If you see Sivaram's answer, he has used the same fact I alluded to you :)