Let a, b, c be integers with at most one of the three equal to 0. Prove that if gcd(a,c)=1, then gcd(ab,c)=gcd(b,c). So far I have two strategies in mind for proving this:
1) Bezout’s Identity: 
ax+cy=1 --> bax+bcy=b ==> gcd(ab,bc)=b
2)Properties of Prime numbers: if gcd(a,c)=1, then a and c do not share any prime factors 
Both ideas have left me at a dead end 
 A: One place to start answering this is, given that $\gcd(a,c)=1$ what can you say about the common factors of $ab$ and $c$.
Try proving the statement that $\gcd(a,c) = 1$ and $a|bc$ implies that $a|b$. Proving this statement should put you on the right track for finding the solution to this problem. 
A: 1) There exist $x,y$ so that $ax + cy=1$ and there are $k,j$ so that $bk + cj=\gcd(b,c)$ so $1*(bk+cj)= (ax + cy)(bk+cj)=abxk + c(ybk+ajx+cyj) =   \gcd(b,c)$.
So $\gcd(ab,c) |\gcd(b,c)$ and clearly $\gcd(b,c)|\gcd(ab,c)$ so $\gcd(ab,c)=\gcd(b,c)$
2) $a$ and $c$ don't share any prime factors.  SO the only prime factor $c$ shares with $ba$ are the prime factors it shares with $b$ so $\gcd(ba,c)=\gcd(b,c)$.
More formally.  Let $\gcd(b,c) = d$ and $b=b'd$ and $c = c'd$.  Then it's easy and standard to show that $c',b'$ are relatively prime [$\gcd(b',c')*d$ is a common divisor of $b$ and $c$ but $d$ is the greatest common divisor].
You should have proven somewhere that $\gcd(ma,mb)= m\gcd(a,b)$.  [($m\gcd(a,b)$ is a common divisor clearly, yet any larger divisor would require taking a common factor from $\frac a{\gcd(a,b)}$ and $\frac b{\gcd(a,b)}$ and there are no common factors left.]
So $\gcd(ab, c) = \gcd(ab'd,c'd) = d\gcd(ab',c')= d*1$.
