China Western Mathematical Olympiad

First, we prove that $a$ is even. It follows from the given condition by choosing an even integer $n \ge 6$.

Next, we prove that $a$ has no odd prime factor. Suppose the contrary, let $p$ be an odd prime factor of $a$. If $p = 3$, let $n = 8$, then $2^n - n^2 = 192$ has a factor $3$, but $a^n - n^a$ is not divisible by $3$, contradicting to $(2^n - n^2) \mid (a^n - n^a)$, hence $p$ is not $3$.

If $p = 5$, let $n = 16$, then $2^n - n^2 = 64 \cdot 110$ has a factor $5$, but $a^n - n^a$ is not divisible by $5$, contradicting $(2^n - n^2) \mid (a^n - n^a)$, hence $p$ is not $5$.

If $p \ge 7$, let $n = p - 1$, it follows from Fermat's Little Theorem that $2^{p-1} \equiv 1 \pmod p$. As $(p-1)^2 \equiv 1 \pmod p$, so $p \mid (2^n - n^2)$. Moreover, since $a$ is even and $p \mid a$, so $n^a \equiv (p-1)^a \equiv (-1)^a \equiv 1 \pmod p$ and $p \mid a^n$, and hence $p$ does not divide $(a^n - n^a)$, contradicting $(2^n - n^2) \mid (a^n - n^a)$.

Finally, we prove that $a$ is $2$ or $4$. For this, let $a = 2^t$ where $t$ is a positive integer, then it follows from $(2^n - n^2) \mid (2^{2n} - n^{2t})$ and $(2^n - n^2) \mid (2^{2n} - n^{2t})$ that $(2^n - n^2) \mid (n^{2t} - n^{2t})$.

If we choose $n$ to be sufficiently large, it follows from the fact $\lim_{n \to \infty} \frac{n^{2t}}{2^n} = 0$ that $n^{2t} - n^{2t} = 0$, hence $2^t = 2t$. $t = 1$ and $t = 2$ are obvious solutions.

If $t \ge 3$, then by the Binomial Theorem, we have $t = 2^{t-1} = (1+1)^{t-1} > 1 + (t-1) = t$, which is impossible. At last, one can easily check that $a = 2$ and $a = 4$ satisfy the condition in the problem, so the solutions for $a$ are $2$ and $4$.