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I searched for natural numbers $n$, where $2^n-n$ and $2^n+n$ are both prime for the range of $n \le 10^5$ on PARI/GP and found that 3, 9 are the only solutions in this range.

Note that since $2^n-n$, $2^n$, $2^n+n$ are in arithmetic progression and $2^n$ is not divisible by 3, $n$ must be a multiple of 3.

Questions:

$(1)$ Are $3$, $9$ the only natural numbers $n$ for which both $2^n-n$ and $2^n+n$ are prime?

$(2)$ If not, then can you prove or disprove that there are finite natural numbers of this type?

Mathphile
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    guess mod 5 was too complicated. –  May 08 '19 at 23:41
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    Ignoring small factors, the chance that either is prime is about$1/\ln (2^n)=1/(n\ln 2)$. The chance that both are prime is the square of that, or $2.08/n^2$. This has a finite sum, so I expect a finite number of solutions. – Empy2 May 09 '19 at 11:16
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    Since you have checked up to $10^5$, and the tail of my sum is $\sum_{n=10^5}^\infty 2.08/n^2\approx 2/10^5$, I would put the chances of any more solutions as $0.00002$ – Empy2 May 09 '19 at 11:26
  • but 31% of valid n values restricted to by 2, and 3 have small factors ( <13). –  May 09 '19 at 11:29
  • @Empy2 how do you know that the chance for finding either prime is $1/ \ln (2^n)$ ? – Mathphile May 09 '19 at 11:30
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    The Prime Number Theorem says there are about $x/\ln x$ primes less than $x$, so the chance of a random $y$ being prime is about $1/\ln y$ – Empy2 May 09 '19 at 11:34
  • turn out I forgot to account for mod 2 losses, so over 62% have small factors. –  May 09 '19 at 12:53
  • There are $30\ 879$ candidates (both $2^n-n$ and $2^n+n$ have no prime factor below $10^5$) in the range $[10^5,10^7]$ – Peter Sep 16 '19 at 08:16
  • Upto $n=250\ 000$ , there is no other pair. – Peter Sep 16 '19 at 21:28

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