Proving $(x, 6) = (x, 8) = 1 \Rightarrow (x, 24) = 1$

Question

Hey I was given this question in my discrete math class, and I'm unsure of what I should do!

Prove that if $x$ is coprime with $6$ and $x$ is coprime with $8$, then $x$ is coprime with 24.

I think I have to use the GCD theorem or co-primality theorem but I don't think what I'm doing is correct but this is what I have so far $$ 1 = ax + by\\ 1 \times 1 = (ax + cy) (bz + cw)\\ \gcd(a, c) = 1\\ \gcd(b, c) = 1\\ \gcd((ab)/2, c) = 1 $$ Thanks in advance!

Answer 1

By Bezout we have \begin{eqnarray*} Ax+6B=1 \\ Cx+8D=1. \end{eqnarray*} Multiply these equations \begin{eqnarray*} x(ACx+6BC+8AD)+24\times 2BD =1. \end{eqnarray*}

Answer 2

Hint: no need to use Bezout, use the contrapositive.

• Suppose that $x$ is not coprime with $24$.
• What can you say about prime factors of $x$?
• Is it possible for $x$ to be coprime with $6$?

Good luck!