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.