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Old 2019-11-06, 22:11   #675
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Nov 2016

6F916 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. there is no finite set {p_1, p_2, p_3, ..., p_u} (all p_i (1<=i<=u) are primes) and finite set {r_1, r_2, r_3, ..., r_s} (all r_i (1<=i<=s) are integers > 1) such that for every integer n>=1:


(k*b^n+c)/gcd(k+c, b-1) is divisible by at least one p_i (1<=i<=u)


k*b^n and -c are both r_i-th powers for at least one r_i (1<=i<=s)


one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t

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

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 2019-11-28 at 03:56
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