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 $2^{p-1}\equiv 1(\mathrm{mod}p)$. As $(p-1{)}^2\equiv 1(\mathrm{mod}p)$, so $p\mid (2^n-n^2)$. Moreover, since $a$ is even and $p\mid a$, so $n^a\equiv (p-1{)}^a\equiv (-1{)}^a\equiv 1(\mathrm{mod}p)$ 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 $4$.