Let $a,b,c,d$ be integers . Suppose that a$ = 3b+2c$ and that $b$ is odd.
Suppose that $b$ is coprime to $c$. Prove that a is coprime to $b$.
so far....
since $b$ is coprime to $c$ there exists integers $x,y$ such that: $1=xb+yc$ by Bezuts lemma.
not sure where to go from to show that: $1= xb+az$ , where $x,z$ are integers.
We try to link $1 = xb + yc$ to the given, $a = 3b+2c$.
Multiplying by $2$ we have:
$$2 = 2xb+ 2yc = 2xb + y(a-3b) = ya + (2x-3y)b$$
By Bezout's Lemma, $\gcd (a,b)$ divides $2$.
Now we just need to prove that $\gcd(a,b) \ne 2$. This can be shown by a condition given by the question that we haven't used yet.
$b$ is odd so $$b=2n+1$$ and b is not divisible by 2.
