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Old 2019-11-06, 22:11   #675
sweety439's Avatar
Nov 2016

22·7·103 Posts

Like Bunyakovsky conjecture, it is conjectured that for all integer triples (k, b, c) satisfying these conditions:

1. k>=1, b>=2, c != 0

2. gcd(k, c) = 1, gcd(b, c) = 1

3. (k*b^n+c)/gcd(k+c, b-1) cannot be proved as composite for all n or prime only for very small n (like the case of (4^n-1)/3) in these three ways:

** Periodic sequence p of prime divisors with p(n) | (k*b^n+c)/gcd(k+c, b-1)
** Algebraic factors (e.g. difference-of-squares factorization, difference-of-cubes factorization, sum-of-cubes factorization, difference-of-5th-powers factorization, sum-of-5th-powers factorization, Aurifeuillian factorization of x^4+4*y^4, etc.) of k*b^n+c
** The combine of them (like the case of 25*12^n-1)

4. the triple (k, b, c) is not in this case: c = 1, b = q^m, k = q^r, where q is an integer not of the form t^s with odd s > 1, and m and r are integers having no common odd prime factor, and the exponent of highest power of 2 dividing r >= the exponent of highest power of 2 dividing m, and the equation 2^x = r (mod m) has no solution (like the case of 8*128^n+1)

Then there are infinitely many integers n>=1 such that (k*b^n+c)/gcd(k+c, b-1) is prime.

Last fiddled with by sweety439 on 2021-04-05 at 16:04
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