Number theory problem (my solution included). Let n be a natural number greater than 1. We call a natural number $a>2$ ,$n-$decomposable if $a^n-2^n$ is divisible by all numbers of the form $a^d + 2^d$, where $d|n$ and $1\le d<n$. Find all $n$ for which there is an $n$-decomposable number.
I got:
$n$ is decomposable $\Leftrightarrow $ n is exactly the form $2^p$.
for "<=", divisors of $n=2^p$ is of form $d=2^q$ where $p\ge q$. since $a^n-2^n=(a-2)(a^1+2^1)(a^2+2^2)(a^4+2^4)(a^8+2^8)\dots(a^n+2^n)$, every $a^d+2^d$ is thereby a factor of $a^n-2^n$
for "=>", by contrapositive, assume $n$ is not $2^p$, then exist odd e and integer $d$ such that $n=de$. Let $b=a^d$ and hence $a^n=b^e$ , then $\frac{a^n-2^n}{a^d+2^d}=\frac{b^e-2^n}{b+2^d}$. by remainder theorem w.r.t. b, the "remainder" is $(-2^d)^e-2^n=-2^n-2^n=-2^{n+1}$.
Now we prove $-2^{n+1}$ is not divisible by $(b+2^d)$. Consider $a=3, b+2^d$ is odd, but $-2^{n+1}$ has only 1 odd factor, namely 1. So $n$ is not decomposable.
 A: The latter part of your proof doesn’t work - it only shows that $3$ is not $n$-decomposible (or by extension, odd numbers are not $n$-decomposible.) It doesn’t answer whether there is any $n$-decomposible number.
And, indeed, there are other $n.$
If $n=p$ is an odd prime, then the only condition is $a+2$ being a factor of $a^p-2^p.$
So $a\equiv-2\pmod {a+2},$ in which case, by your argument, $$a^{p}-2^{p}\equiv -2^{p+1}\pmod {a+2}.$$
So we only need $a+2\mid 2^{p+1}.$ So $a=2^k-2$ for some $3\leq k\leq p+1.$
For example, $a=2^3-2=6$ is always $p$-decomposible for all odd primes $p$ because $6+2=8$ and $6^p-2^p$ is divisible by $8.$
$a=2^5-2=30$ is $p$-decomposible for prime $p\geq 5.$
It might be the case that $n$ prime or $n>1$ a power of $2$ are all the cases, I don’t know. Certainly, it seems harder to find $a$ when $n$ has a lot of odd factors.
When $n$ is even and not a power of $2,$ showing there are no $n$-reducible numbers reduces to the case of solving $m^2+1=2^k,$ which only has solutions $m=0,1,$ by looking modulo $4.$
So you are left with $n$ odd and not prime. Your proof again shows that $a=2b$ must be even. Then you need $b^d+1$ is a power of $2$ less than or equal to $2^{n+1-d}$ for all  factor $d<n.$ That seems unlikely for even $b+1$ and $b^p+1,$ but I don’t have the proof.
It certainly reduces the possible $a,$ since if $n>1$ is odd and not prime and $p$ is the is the smallest fact or $n,$ you get $b^{n/p}\leq 2^{n+p}$ or $b\leq 2^p\cdot 2^{p^2/n}\leq2^{1+\sqrt{n}}.$
