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.