Prime divisors of numbers 2^n + 3 I'm interested in the following problem: do there exist infinitely many prime numbers $p$ such that $p^2|2^{n}+3$ for some natural number $n$?
Some motivation:
If we replace the function $2^n + 3$ with the $f(n)$ where $f \in \mathbb{Z}[x]$ is non-constant that this is true (follows Hensel lemma).
So, it's rather natural to try proving this for some other non-polynomial functions. $2^n + 3$ is an easy example of such function. There is also another good reason: sequence $a_n = 2^{n} + 3$ satisfies the reccurence relation: $a_{n+2} = 3a_{n+1} - 2a_{n}$. And for example this problem is true for Fibonacci sequence. So, for Fibonacci it's easier even if the closed form of Fibonacci numbers is more complicated. But I think that the reason of this is that the Fibonacci numbers satisfy some "good" identities which other sequences don't have to share.
Now some remarks:
It's easy exercise to prove that there are infinitely many primes $p$ such that $p|2^{n}+3$. Also, if we try "correcting" $n$ to work also for $p^2$ and we try $m=n+k(p-1)$ we see that it is possible unless $p$ is Wieferich prime, i.e. satisfies $p^2|2^{p-1}-1$. And this gives us nothing as we don't know much about Wieferich primes...
This method can of course be generalized in such way: if $p|2^{n}+3$ and order of $2$ mod $p^2$ is greater than order of $2$ mod $p$ then we can find $m$ such that $p^2|2^{m}+3$. But I don't really think that it helps.
I'm interested in some information about this problem (especially if it's open or not) and also related problems. We can ask a general question: for which functions $f$ we know that this is true?
Edit: Sorry for the confusion with $k$, deleted.
 A: (Edited as the comments below suggest)
The ABC conjecture seemed to me like it would play a roll, however it comes up a little short:
"Are there infinitely many primes $p$ so that for each $p$ there is some integer $n$ with $p^2|2^n + 3?"$
If the ABC conjecture is true, then this answer to this question is almost "no", but still there is a problem at the end of the argument.
The ABC conjecture states that for any $\epsilon > 0$ there is a constant $K_\epsilon$ so that for any co-prime triple $A < B < C$ with $A+B = C$ then 
$$C \le K_\epsilon\prod_{p|ABC}p^{1 + \epsilon}.$$
So, if there is such an infinite collection of primes, then for the corresponding infinite $n$ where this is true then $2^n + 3 = p^2C$ then
$$p^2C \le K_\epsilon(6Cp)^{1+\epsilon}.$$ 
(Edited: The following sentence is incorrect "But this will clearly run into problems for sufficiently large $p.$" But I wanted to leave it so Kevin's comment makes sense.)
Note that as $C = C(p)$ is a function of $p$ then the $C^\epsilon$ (when $C$ is square-free, or nearly square-free) term may still allow this inequality to work.
A: This is part of the question I asked on 5191. On my observation primes categorized themselves into 3 types, a)primes that doesn't divide the form of 2^x + c, b)primes that divides but a smaller had divided it already and c)primes that divides a new x usually larger primes. Once a prime divides a form in this case 2^x + 3 it will divide it infinitely periodically, so I believe even the form of p^k|2^x + c will be true. All primes that has a period of p-1 (totient = p-1) will always divide any forms. On the case of wiefirich prime is a special form, all integer divides a mersenne, so no problem with any p^k dividing a mersenne, but tying the order of mersenne 2^k - 1 to prime k, it could just be one of the small numbers properties that the 2 wiefirich primes what they are. 
