Problems & Puzzles: Problems

Problem 52.  p 2^n

F. Cohen and J. L. Selfridge observed in 1975 that 47867742232066880047611079 is a prime number such that plus or minus a power of 2 can never be prime.

 

Q1. Can you find a smaller solution?

Q2. Just take p+2^n and get champions of this type

 

 


Arkadiusz Wesoloski wrote on August1, 2017:

A prime Brier number is a prime n such that the numbers n*2^k + 1 and n*2^k - 1 are composite, for all natural numbers k. The first few are 10439679896374780276373, 21444598169181578466233, 105404490005793363299729, 178328409866851219182953, 239365215362656954573813, 378418904967987321998467, 422280395899865397194393, 474362792344501650476113, 490393518369132405769309.

The above appears as A180247 in OEIS.

Each of these numbers is a solution to question 1 of this Problem. Why?

"Let k be an odd number which is simultaneously Sierpinski and Riesel, let A be a set of primes which divide numbers of the form k*2^m + 1 and of the form k*2^m - 1 with m >= 1, and let B be a set of primes which divide numbers of the form k + 2^n and of the form abs(k - 2^n) with n >= 1. Sets A and B are equal; both sets have exact same elements."

...there is no need to create a proof; I used a trick. Observe that:

1) if a prime p is a factor of k*2^n + 1 for some n >= 0 and p > n, then p divides k + 2^(p-n-1).
2) if an odd prime p is a factor of k + 2^n for some n >= 0 and p > n, then p divides k*2^(p-n-1) + 1.
3) if a prime p is a factor of k*2^n - 1 for some n >= 0 and p > n, then p divides abs(2^(p-n-1) - k).
4) if an odd prime p is a factor of 2^n - k for some n >= 0 and p > n, then p divides k*2^(p-n-1) - 1.
These statements are simple consequences of Fermat's little theorem.
 
That's all.

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