View Single Post 2019-11-06, 22:11 #675 sweety439   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  