The problem statement remains the same, and I'll reiterate the proof below for clarity.

Problem:

We are required to prove that if: $\gcd(a, m) = 1 \quad \text{and} \quad \gcd(b, m) = 1,$ then: $\gcd(ab, m) = 1.$

Proof:

1. Given:

   • $\gcd(a, m) = 1$: This means that $a$ and $m$ have no common prime factor other than 1.
   • $\gcd(b, m) = 1$: Similarly, $b$ and $m$ have no common prime factor other than 1.

2. To Prove:

   • $\gcd(ab, m) = 1$: The product $ab$ and $m$ have no common prime factor other than 1.

3. Prime Factorization:

   • Let the prime factorization of $m$ be $p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$.
   • Since $\gcd(a, m) = 1$, none of the primes $p_1, p_2, \dots, p_k$ divide $a$.
   • Similarly, since $\gcd(b, m) = 1$, none of the primes $p_1, p_2, \dots, p_k$ divide $b$.

4. Product $ab$:

   • Since none of the primes $p_1, p_2, \dots, p_k$ divide either $a$ or $b$, none of them can divide the product $ab$.
   • This implies that $ab$ and $m$ share no common prime factor.

5. Conclusion:

   • Therefore, $\gcd(ab, m) = 1$.

Alternative Argument Using GCD Properties:

The gcd satisfies the property: $\gcd(x, yz) = 1 \iff \gcd(x, y) = 1 \text{ and } \gcd(x, z) = 1.$

• Here, treat $ab$ as $(a \cdot b)$:
  • Since $\gcd(a, m) = 1$ and $\gcd(b, m) = 1$, we have: $\gcd(a \cdot b, m) = 1.$

Thus, $\gcd(ab, m) = 1$.

Final Statement:

It is proven that if $\gcd(a, m) = 1$ and $\gcd(b, m) = 1$, then $\gcd(ab, m) = 1$.

Would you like further explanation or alternative proofs? Here are some related questions to explore:

1. How can we generalize this proof for more numbers (e.g., $a, b, c$)?
2. What is the role of the fundamental theorem of arithmetic in gcd proofs?
3. Can this be proved using modular arithmetic concepts?
4. How does this relate to the coprime property in group theory?
5. What other gcd properties can simplify problems like this?

Tip: When working with gcd, remember to rely on its basic properties, especially for factorization and distribution over multiplication.