The answer is all $n=2^k$, where $k>1$ is arbitrary. Note that any such $n$ clearly works: \[ a^{2^k}-2^{2^k} = (a-2)(a+2)\left(a^2+2^2\right)\left(a^{2^2}+2^{2^2}\right)\cdots \left(a^{2^{k-1}} + 2^{2^{k-1}}\right). \]We now show that if $n$ is not a power of two, then this is impossible. Set $n=2^k\cdot n'$, where $n'>1$ is odd; and $k\ge 0$. Assume first that $k>0$. Then observe that \[ a^{2^k}\equiv -2^{2^k}\pmod{a^{2^k}+2^{2^k}}\implies a^n \equiv -2^n\pmod{a^{2^k}+2^{2^k}}. \]From here, we conclude $a^{2^k}+2^{2^k}\mid 2^{n+1}$. That is, $a^{2^k}+2^{2^k}=2^u$ for some $u$. Clearly $a$ is even, set $a=2a'$, and $a'>1$. We then have \[ 2^{2^k}\left((a')^{2^k}+1\right) = 2^u \implies (a')^{2^k}+1=2^t\quad\text{for some}\quad t\ge 2. \]However, as $k>0$ and $t\ge 2$ (since $a'>1$), this is impossible via modulo $4$. Likewise, if $k=0$ (that is $n$ is odd), set $n=pq$ for some $p,q>1$ odd. We have \[ a^p\equiv -2^p\pmod{a^p+2^p}\implies a^n\equiv -2^n\pmod{a^p+2^p}\implies a^p+2^p = 2^u \quad\text{for some }\quad u. \]Again, setting $a=2a’$ for some $a’>0$, we find $(a’)^p+1$ to be a power of two. Clearly, $p>1$ and $(a’)^{p-1}-(a’)^{p-2}+\cdots+1$ is a factor of this object, which is odd and is greater than one; yielding a contradiction.

grupyorum wrote: The answer is all $n=2^k$, where $k>1$ is arbitrary. Note that any such $n$ clearly works: \[ a^{2^k}-2^{2^k} = (a-2)(a+2)\left(a^2+2^2\right)\left(a^{2^2}+2^{2^2}\right)\cdots \left(a^{2^{k-1}} + 2^{2^{k-1}}\right). \]We now show that if $n$ is not a power of two, then this is impossible. Set $n=2^k\cdot n'$, where $n'>1$ is odd; and $k\ge 0$. Assume first that $k>0$. Then observe that \[ a^{2^k}\equiv -2^{2^k}\pmod{a^{2^k}+2^{2^k}}\implies a^n \equiv -2^n\pmod{a^{2^k}+2^{2^k}}. \]From here, we conclude $a^{2^k}+2^{2^k}\mid 2^{n+1}$. That is, $a^{2^k}+2^{2^k}=2^u$ for some $u$. Clearly $a$ is even, set $a=2a'$, and $a'>1$. We then have \[ 2^{2^k}\left((a')^{2^k}+1\right) = 2^u \implies (a')^{2^k}+1=2^t\quad\text{for some}\quad t\ge 2. \]However, as $k>0$ and $t\ge 2$ (since $a'>1$), this is impossible via modulo $4$. Likewise, if $k=0$ (that is $n$ is odd), set $n=pq$ for some $p,q>1$ odd. We have \[ a^p\equiv -2^p\pmod{a^p+2^p}\implies a^n\equiv -2^n\pmod{a^p+2^p}\implies a^p+2^p = 2^u \quad\text{for some }\quad u. \]Again, setting $a=2a’$ for some $a’>0$, we find $(a’)^p+1$ to be a power of two. Clearly, $p>1$ and $(a’)^{p-1}-(a’)^{p-2}+\cdots+1$ is a factor of this object, which is odd and is greater than one; yielding a contradiction. Something wrong in this solution, can you explain me why $$\begin{cases} a^{2^k}\equiv -2^{2^k}\pmod {a^{2^k}+2^{2^k}}\\ a^n\equiv -2^n\pmod {a^{2^k}+2^{2^k}} \end{cases}\implies a^{2^k}+2^{2^k}\mid 2^{n+1}$$Similarly with the case $n=pq$ for some $p,q>1$ odd.

Nothing wrong here. Clearly $a^{2^k}+2^{2^k}$ divides $a^{2^kn'}-2^{2^kn'}=a^n+2^n$ since $n$ is odd. Since it also divides $a^n-2^n$, it divides their difference which is $2^{n+1}$.

Tintarn wrote: Nothing wrong here. Clearly $a^{2^k}+2^{2^k}$ divides $a^{2^kn'}-2^{2^kn'}=a^n+2^n$ since $n$ is odd. Since it also divides $a^n-2^n$, it divides their difference which is $2^{n+1}$. Thank you very much

Thus $n=2^b \cdot c$ where $c$ is odd, because of divisibility conditions $a^{2^b} + 2^{2^b} \mid 2^{n+1} \implies a^{2^b} + 2^{2^b}=2^e$ where $e > 2^b$ and basically following the same proof style in (preliminary result) this is not possible. Clearly when $n$ is a power of two, it works and is the only solution.

The answer is $n=2^k$, where $k>1$ is a positive integer. To show it works, $a^{2^k} \equiv (a^d)^{\frac{2^k}{d}} \equiv (-2^d)^{\frac{2^k}{d}} \equiv 2^{2^k}$, where $\frac{2^k}{d}$ is obviusly even. Suppose $n$ is not a power of two. Then $n=pk$, where $p>2$ is a prime. Substitute $d=k$ to get: $$2^n \equiv a^n \equiv (a^k)^p \equiv (-2^k)^p \equiv -2^{kp} \equiv -2^n$$Thus $2^{n+1} \vdots a^d+2^d$, so $a^d+2^d$ is a power of two for every $d | n, d \neq n$. For $d=1$: $a+2=2^f$, hence $a=2^f-2$. Then $2^d((2^{f-1}-1)^d+1)$ is a power of two. But then $d=p$ breaks by LTE lemma. Contradiction.

Solved with stillwater_25. We claim that the answer is all positive integers of the form $n=2^k$ for some $k >1$. We first note that such $n$ do work since, \[a^{2^{k}}-2^{2^{k}}=(a-2)(a+2)(a^2+2^2)\dots (a^{2^{k-1}}+2^{2^{k-1}})\]Since any divisor $d$ of $n$ is a power of two less than $2^k$ the result follows. Now, note that if $n$ is not a power of two, $d=2^{\nu_2(n)}<n$ is a proper divisor of $n$. Further, $\frac{n}{d}$ will be an odd positive integer. Thus, \[a^d + 2^d \mid a^n + 2^n\]We are also given that, \[a^d + 2^d \mid a^n - 2^n\]Summing we obtain, \[a^d + 2^d \mid 2^{n+1}\]which implies that $a^d+2^d=2^i$ for some positive integer $i$. It is not hard to check that this equation has solutions only if $a=2$ and $i=d+1$. But since we are given that $a>2$ this is impossible, and we are done.