# proof of property of gcd

• Jul 10th 2011, 09:54 PM
oblixps
proof of property of gcd
prove or disprove the following fact: if gcd(b,c) = 1 then gcd(ab,c) = gcd(a, bc).

i am trying to show that if a number is a common divisor to both a and bc then it must be a common divisor to ab and c. using the fact that bx + cy = 1 for some x and y, i find that if a number is a common divisor to both a and bc then it must be a common divisor to ab as well from the fact that (b)(b)(x)a + (y)(a)bc = ab. what i am having trouble showing is that this number must divide c too.

i tried a proof by contradiction by saying that this number does not divide c but i couldn't find a contradiction. maybe this fact is untrue? if so, then would the best way to disprove it to find a counterexample? thanks
• Jul 10th 2011, 10:20 PM
melese
Re: proof of property of gcd
You can show (using Euclid's Lemma) that if $gcd(b,c)=1$, then $gcd(ab,c)=gcd(a,c)$.

Counterexample to the original statement: take $a=4, b=6, c=5$.