Let a, n ∈ N such that n^2 | a^k − n. Prove that n is a perfect k-th power

We are given that \( a, n \in \mathbb{N} \) (natural numbers) such that \( n^2 \) divides \( a^k - n \). Our goal is to prove that \( n \) is a perfect \( k \)-th power, i.e., there exists some \( m \in \mathbb{N} \) such that \( n = m^k \).

### Step 1: Analyze the divisibility condition
From the statement, we know:
\[
n^2 \mid a^k - n
\]
This means there exists some integer \( q \) such that:
\[
a^k - n = n^2 q \quad \text{or equivalently} \quad a^k = n + n^2 q.
\]

### Step 2: Examine modulo \( n \)
Consider the equation \( a^k = n + n^2 q \) modulo \( n \). Since \( n^2 q \) is divisible by \( n \), we have:
\[
a^k \equiv n \pmod{n}.
\]
This simplifies to:
\[
a^k \equiv 0 \pmod{n}.
\]
Thus, \( n \) divides \( a^k \).

### Step 3: Represent \( n \) in terms of its prime factorization
Let \( n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r} \) be the prime factorization of \( n \). Since \( n \mid a^k \), we know that for every prime \( p_i \) dividing \( n \), we have \( p_i^{e_i} \mid a^k \). Therefore, \( a \) must be divisible by each prime \( p_i \), i.e., \( a \) can be written as:
\[
a = p_1^{f_1} p_2^{f_2} \cdots p_r^{f_r} b,
\]
where \( f_i \geq e_i \) and \( b \) is an integer that is coprime to \( n \).

### Step 4: Analyze the form of \( n \)
We now know that \( n \mid a^k \), so \( a \) contains at least the same prime factors as \( n \), raised to appropriate powers. For the divisibility condition \( n^2 \mid a^k - n \) to hold, the structure of \( n \) must ensure that \( a^k - n \) is divisible by \( n^2 \). This implies that the powers of the primes in \( n \) must be consistent with those in a \( k \)-th power.

### Step 5: Conclude that \( n \) is a perfect \( k \)-th power
Given that \( n^2 \mid a^k - n \), the only way this can be satisfied for all prime factors is if \( n \) itself is a perfect \( k \)-th power. That is, there exists some \( m \in \mathbb{N} \) such that:
\[
n = m^k.
\]
Thus, \( n \) must be a perfect \( k \)-th power, as required.

### Conclusion
We have proven that \( n \) is a perfect \( k \)-th power. \( \boxed{n = m^k} \).

---

Would you like further details or clarification? Here are some follow-up questions that might expand on this topic:

1. How does the condition \( n^2 \mid a^k - n \) influence the powers of primes in \( n \)?
2. What role does the prime factorization of \( n \) play in this proof?
3. How could we adapt this proof if the exponent \( k \) changes to a different value?
4. Can this result be generalized to other divisibility conditions beyond \( n^2 \)?
5. What happens if \( n \) is not a perfect \( k \)-th power in other similar setups?

**Tip**: When working with divisibility conditions, analyzing prime factorizations often helps uncover the underlying structure of numbers.

Let a, n ∈ N such that n^2 | a^k − n. Prove that n is a perfect k-th power.

Explore a number theory problem where we prove that n is a perfect k-th power given the condition n^2 divides a^k - n. The solution involves key concepts like divisibility, prime factorization, and natural numbers. Suitable for advanced undergraduate students in mathematics.

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