48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Let the prime factorization of $b$ be $b=p_1^{{\alpha }_1}\dots p_s^{{\alpha }_s}$, where $p_1,\dots ,p_s$ are distinct primes. Our goal is to show that all exponents ${\alpha }_i$ are divisible by $n$, then we can set $A=p_1^{{\alpha }_1/n}\dots p_s^{{\alpha }_s/n}$.

Apply the condition for $k=b^2$. The number $b-a_k^n$ is divisible by $b^2$ and hence, for each $1\le i\le s$, it is divisible by $p_i^{2{\alpha }_i}>p_i^{{\alpha }_i}$ as well. Therefore $$a_k^n\equiv b\equiv 0(\mathrm{mod}{p}_i^{{\alpha }_i})$$ and $$a_k^n\equiv b\not\equiv 0(\mathrm{mod}{p}_i^{{\alpha }_i+1})$$ 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 $n$.