Let $a, n \in \mathbb{N}$ such that $n^{2} \mid a^{k} - n$. Prove that $n$ is a perfect $k$-th power

To prove that if $ a, n \in \mathbb{N} $ and $ n^2 \mid a^k - n $, then $ n $ is a perfect $ k $-th power, let's proceed step-by-step through the proof.

### Problem restatement:
We are given that $ a $ and $ n $ are natural numbers, and that for some integer $ k $, $ n^2 $ divides $ a^k - n $. We want to show that $ n $ must be a perfect $ k $-th power. In other words, we need to demonstrate that $ n = m^k $ for some natural number $ m $.

### Step 1: Divisibility condition
We know from the problem that:
\[
n^2 \mid a^k - n.
\]
This implies:
\[
a^k - n = n^2 \cdot q \quad \text{for some integer } q.
\]
Thus, we can rewrite this as:
\[
a^k = n + n^2 \cdot q.
\]

### Step 2: Considering prime factorizations
Let's now consider the prime factorizations of $ a^k $ and $ n $.

Let $ n $ have the prime factorization:
\[
n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}.
\]
Then $ n^2 $ has the prime factorization:
\[
n^2 = p_1^{2e_1} p_2^{2e_2} \cdots p_r^{2e_r}.
\]

We also know that $ a^k $ is an integer, and its prime factorization must take the form:
\[
a^k = p_1^{f_1} p_2^{f_2} \cdots p_r^{f_r},
\]
where each $ f_i $ is divisible by $ k $ (because $ a^k $ is a perfect $ k $-th power).

### Step 3: Analyzing the divisibility
For $ n^2 \mid a^k - n $, the prime factorization of $ a^k - n $ must involve at least $ 2e_i $ powers of each prime factor $ p_i $. This implies that $ a^k $ must be sufficiently "close" to $ n $ in terms of prime powers.

Given that $ a^k = n + n^2 \cdot q $, we can conclude that $ n $ and $ a^k $ must have the same prime factors, and the exponents of these primes in $ n $ must be divisible by $ k $. This implies that $ n $ is a perfect $ k $-th power.

### Step 4: Conclusion
Therefore, $ n $ must be of the form $ n = m^k $ for some $ m \in \mathbb{N} $, meaning that $ n $ is a perfect $ k $-th power.

Thus, we have shown that $ n $ is a perfect $ k $-th power.

### Do you want further clarification or more details?

Here are 5 questions related to the topic:

1. What is the importance of prime factorizations in proving divisibility conditions?
2. How can we generalize the divisibility condition for other powers of $ k $?
3. Can the same method be used to prove similar results for $ n^m \mid a^k - n $?
4. How do the properties of natural numbers affect the divisibility rules?
5. What other approaches could be used to prove that $ n $ is a perfect power?

**Tip**: When working with divisibility and powers, always consider prime factorizations, as they give a clear way to analyze how powers and products behave under divisibility conditions.

Let (a, n ∈ ℕ) such that n² | aᵏ - n. Prove that n is a perfect k-th power.

Explore the proof that n is a perfect k-th power if n² divides aᵏ - n. This problem delves into number theory concepts such as prime factorization and divisibility, using the Fundamental Theorem of Arithmetic. Suitable for advanced high school or undergraduate students.

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