, 2007

Let the prime factorization of $b$ be $b = p_{1}^{\alpha_{1}} \ldots p_{s}^{\alpha_{s}}$, where $p_{1}, \ldots, p_{s}$ are distinct primes. Our goal is to show that all exponents $\alpha_{i}$ are divisible by $n$, then we can set $A = p_{1}^{\alpha_{1} / n} \ldots p_{s}^{\alpha_{s} / n}$.

Apply the condition for $k = b^{2}$. The number $b - a_{k}^{n}$ is divisible by $b^{2}$ and hence, for each $1 \leq i \leq s$, it is divisible by $p_{i}^{2 \alpha_{i}} > p_{i}^{\alpha_{i}}$ as well. Therefore
$$a_{k}^{n} \equiv b \equiv 0 \quad (\bmod p_{i}^{\alpha_{i}})$$
and
$$a_{k}^{n} \equiv b \not\equiv 0 \quad (\bmod p_{i}^{\alpha_{i} + 1})$$
which implies that the largest power of $p_{i}$ dividing $a_{k}^{n}$ is $p_{i}^{\alpha_{i}}$. Since $a_{k}^{n}$ is a complete $n$th power, this implies that $\alpha_{i}$ is divisible by $n$.