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3 added 66 characters in body

(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 noalmost "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.

2 added 235 characters in body

I'm a little confused by your first question, because I cannot see where the $k$ comes in

(Edited as Theo pointed out above). I'll answer the possible question which you may be askingcomments 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 no.

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 $p.$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$is square-free, or nearly square-free) term may still allow this inequality to work. 1 I'm a little confused by your first question, because I cannot see where the$k$comes in (as Theo pointed out above). I'll answer the possible question which you may be asking: "Are there infinitely many primes$p$so that for each$p$there is some integer$n$with$p^2|2^n + 3?"$If the ABC conjecture is true, then this answer to this question is no. The ABC conjecture states that for any$\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 $$p^2C \le K_\epsilon(6Cp)^{1+\epsilon}.$$ But this will clearly run into problems for sufficiently large$p.\$