Nice problem!

Let $p \in \mathbb{P}$ such that $p|a$. If possible, suppose that $p\ge 3$. Let $n=p-1$. Therefore, $p|2^{p-1}-(p-1)^2|a^{p-1}-(p-1)^a$. But, $p|a \implies p|(p-1)^a$, a clear contradiction. As a result, we infer that either $a=1$ or $a=2^k$. Case I: $a=1 \implies 2^n-n^2|1-n \implies 2^n-n^2|n-1 \forall n \in \mathbb{N}$, but $\forall n \ge 5$, $2^n-n^2 >n-1$, contradiction. Case II: $a=2^k \implies 2^n-n^2|2^{kn}-n^{2^k}$. Since $a-b|a^m-b^m \forall m \in \mathbb{N}$, we have \[2^n-n^2|2^{kn}-n^{2k}\]. \[ 2^n-n^2|n^{2^k}-n^{2k} \] If $k=2$ or $1$, this is obvious. Let $n = 2^{2^k} \implies 2^{2^{2^k}}-2^{2^{k+1}} \le 2^{2^{2k}}-2^{k.2^{k+1}}$, and by induction, this clearly impossible for $k \ge 3$. Therefore, $a=2$ or $a=4$.

Very Nice!

The 2013 China West Mathematics Invitation Competition was held in the high school affiliated to northwest normal university from 15th Aug. to 19th Aug with 21 representative teams including 105 teenager talents for mathematics taking part in, who come from different regions or countries covering 14 provinces or cities such as Beijing, Chongqing, Shaanxi, Gansu, Guangxi, Guizhou, Hainan, Jiangxi, Ningxia, Qinghai, Sichuan, Xinjiang, Yunnan, etc. and Hong Kong Special Administrative Region as well as some foreign countries like Singapore, Kazakhstan, Indonesia and so on. The Examiners Committee of this competition consists of several professors or experts from different universities or high schools such as Shanghai University, Peking University, University of Macau, Lanzhou University, East China Normal University, and the High School Attached to Renmin University of China, the High School Attached to Hunan Normal University and so forth.

There is a better proof that a friend of mine found at the actual competition.

DVDthe1st wrote: There is a better proof that a friend of mine found at the actual competition.

Wow!! It is really nice !! Thank you very much.

Good to know that Chineses used my problem http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=495826

sqing wrote: Find all positive integers $a$ such that for any positive integer $n\ge 5$ we have $2^n-n^2\mid a^n-n^a$. $2^n-n^2\mid a^n-n^a$. $\therefore 2^n-n^2\mid a^{2n}-n^{2a}$. But $2^n-n^2\mid 2^{na}-n^{2a}$. So $2^n-n^2\mid a^{2n}-2^{na}$. Choose prime $p$ s.t. $p$ is greater than $a^2$ and $2^a$. $n=p-1\implies 2^{p-1}-(p-1)^2\mid a^{2(p-1)}-2^{(p-1)a}$. Fermat's little theorem $\implies p\mid 2^a-a^2\implies 2^a-a^2=0$ $\implies a =$ $2$ or $4$. Checking we find that $(2,4)$ indeed satisfy the condition.

Let $p>|2^a-a^2|$ be a prime and $n=p(p-1)+2.$ Then $2^n=(2^{p-1})^p\cdot 2^2\equiv 2^2\pmod p,n^2\equiv 2^2\pmod p.$ Therefore $p\mid 2^n-n^2\Rightarrow p\mid a^n-n^a.$ However $a^n=(a^{p-1})^p\cdot a^2\equiv a^2\pmod p,n^a\equiv 2^a\pmod p.$ Then $0\equiv a^n-n^a\equiv 2^a-a^2\pmod p.$ Using $|2^a-a^2|<p,2^a-a^2=0,a=2$ or ${4}{}$ which is both true$.\blacksquare$

Taking $n$ even, we get that $a$ is even, so $a=2k$ $2^n-n^2 | (2k)^n-(2^k)^n$. Now if $p | 2^n-n^2$, then $(2^k)^n \equiv (2k)^n (mod p)$, so either $(n,p-1)>1$, or $2^k-2k \vdots p$. Now take any big prime $p$ such that $2$ is a quadratic residue mod $p$. By Chinese Remainder Theorem there exist $n$ such that $n \equiv \sqrt{2}$ (mod $p$) and $n \equiv 1$ (mod $p-1$). So $p | 2^n-n^2$ and $(n,p-1)=1$, so $2^k-2k \vdots p$. Because we can take $p$ arbitrarily large, $2^{k-1}=k$, so $k=1$ or $2$, so $a=2$ or $4$, both work