To prove that $a \mid b$ if and only if $a^n \mid b^n$, we will prove both directions of the equivalence:

Proof:

1. (If direction): Assume $a \mid b$. Show that $a^n \mid b^n$.

   Since $a \mid b$, there exists an integer $k$ such that: $b = a \cdot k.$

   Raising both sides to the power of $n$: $b^n = (a \cdot k)^n = a^n \cdot k^n.$

   Since $k^n$ is an integer, it follows that $a^n \mid b^n$.

2. (Only if direction): Assume $a^n \mid b^n$. Show that $a \mid b$.

   Since $a^n \mid b^n$, there exists an integer $m$ such that: $b^n = a^n \cdot m.$

   To prove $a \mid b$, let's analyze this equation in terms of prime factorization:

   • Let $a = p_1^{x_1} p_2^{x_2} \cdots p_k^{x_k}$ and $b = p_1^{y_1} p_2^{y_2} \cdots p_k^{y_k}$, where $p_i$ are primes.
   • Then $a^n = p_1^{n x_1} p_2^{n x_2} \cdots p_k^{n x_k}$ and $b^n = p_1^{n y_1} p_2^{n y_2} \cdots p_k^{n y_k}$.

   Since $a^n \mid b^n$, we have: $n x_i \leq n y_i \quad \text{for all } i.$

   Dividing by $n$, we get: $x_i \leq y_i \quad \text{for all } i.$

   This implies $a \mid b$.

Conclusion:

We have shown both directions, thus proving $a \mid b$ if and only if $a^n \mid b^n$.

Do you want any details or have any questions?

Related Questions:

1. What are some examples of divisibility properties with specific values of $a$, $b$, and $n$?
2. How does this proof change if $n = 1$?
3. Can this proof be extended to other mathematical structures, such as rings?
4. How does this concept apply to polynomial divisibility?
5. What are other common divisibility theorems in number theory?

Tip:

Always break down divisibility proofs using definitions and factorization to clearly see the relationships between terms.