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Old 2020-02-16, 19:17   #1
enzocreti
 
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Default Primes of the form 56^n+1

Could be primes of the form
56^n+1 finitely many?
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Old 2020-02-16, 20:19   #2
Batalov
 
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Quote:
Originally Posted by enzocreti View Post
Could be primes of the form
56^n+1 finitely many?
Well known that finitely few.

Sit down with a piece of paper and think what values can n take for 56^n+1 to be prime?
Only powers of 2. b^2^m+1 are called generalized Fermat numbers. Read Mathworld for a starter.
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Old 2020-02-17, 15:56   #3
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Quote:
Originally Posted by enzocreti View Post
Could be primes of the form
56^n+1 finitely many?
See post https://mersenneforum.org/showpost.p...&postcount=675, I conjectured that there are infinitely many integers n>=1 such that (k*b^n+c)/gcd(k+c, b-1) is prime 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:

either

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

or

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

or

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.

** Examples of the condition 3:

(k, b, c) = (78557, 2, 1): (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n: {3, 5, 7, 13, 19, 37, 73}

(k, b, c) = (509203, 2, -1): (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n: {3, 5, 7, 13, 17, 241}

(k, b, c) = (419, 4, 1): (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n: {3, 5, 7, 13}

(k, b, c) = (334, 10, -1): (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n: {3, 7, 13, 37}

(k, b, c) = (7, 5, 1): (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n: {2, 3}

(k, b, c) = (13, 5, -1): (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n: {2, 3}

(k, b, c) = (1, 4, -1): k*b^n and -c are both 2nd powers for all n

(k, b, c) = (9, 4, -1): k*b^n and -c are both 2nd powers for all n

(k, b, c) = (1, 9, -1): k*b^n and -c are both 2nd powers for all n

(k, b, c) = (4, 9, -1): k*b^n and -c are both 2nd powers for all n

(k, b, c) = (1, 8, 1): k*b^n and -c are both 3rd powers for all n

(k, b, c) = (1, 8, -1): k*b^n and -c are both 3rd powers for all n

(k, b, c) = (8, 27, 1): k*b^n and -c are both 3rd powers for all n

(k, b, c) = (8, 27, -1): k*b^n and -c are both 3rd powers for all n

(k, b, c) = (4, 16, -1): "one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t" for all n

(k, b, c) = (4, 81, -1): "one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t" for all n

(k, b, c) = (324, 16, -1): "one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t" for all n

(k, b, c) = (64, 81, -1): "one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t" for all n

(k, b, c) = (4, 19, -1): k*b^n and -c are both 2nd powers for all even n, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all odd n: {5}

(k, b, c) = (4, 24, -1): k*b^n and -c are both 2nd powers for all even n, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all odd n: {5}

(k, b, c) = (1369, 30, -1): k*b^n and -c are both 2nd powers for all even n, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all odd n: {7, 13, 19}

(k, b, c) = (400, 88, -1): k*b^n and -c are both 2nd powers for all even n, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all odd n: {3, 7, 13}

(k, b, c) = (3600, 270, -1): k*b^n and -c are both 2nd powers for all even n, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all odd n: {7, 13, 37}

(k, b, c) = (343, 10, -1): k*b^n and -c are both 3rd powers for all n divisible by 3, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n not divisible by 3: {3, 37}

(k, b, c) = (3511808, 63, 1): k*b^n and -c are both 3rd powers for all n divisible by 3, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n not divisible by 3: {37, 109}

(k, b, c) = (64, 957, -1): k*b^n and -c are both 3rd powers for all n divisible by 3, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n not divisible by 3: {19, 73}

(k, b, c) = (2500, 55, 1): "one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t" for all n divisible by 4, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n not divisible by 4: {7, 17}

(k, b, c) = (16, 200, 1): "one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t" for all n = 2 mod 4, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n not = 2 mod 4: {3, 17}

(k, b, c) = (64, 936, -1): k*b^n and -c are both 2nd powers for all even n, k*b^n and -c are both 3rd powers for all n divisible by 3, (k*b^n+c)/gcd(k+c, b-1) is divisible by at least one primes in this set for all n = 1 or 5 mod 6: {37, 109}

** Examples of the condition 4:

(k, b, c) = (8, 128, 1)

(k, b, c) = (32, 128, 1)

(k, b, c) = (64, 128, 1)

(k, b, c) = (8, 131072, 1)

(k, b, c) = (32, 131072, 1)

Last fiddled with by sweety439 on 2020-02-17 at 15:59
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Old 2020-02-17, 15:58   #4
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Quote:
Originally Posted by enzocreti View Post
Could be primes of the form
56^n+1 finitely many?
In your case is (k,b,c) = (1,56,1), satisfying all four conditions, thus I conjectured that there are infinitely many primes of this form, additionally, I conjectured that there are infinitely many primes of the form (k*56^n+1)/gcd(k+1,56-1) for all k<20 (there are no such primes for k=20 because of the covering set {3,19}, see extended Sierpinski/Riesel problems.
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Old 2020-02-17, 16:13   #5
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Quote:
Originally Posted by Batalov View Post
Well known that finitely few.

Sit down with a piece of paper and think what values can n take for 56^n+1 to be prime?
Only powers of 2. b^2^m+1 are called generalized Fermat numbers. Read Mathworld for a starter.
Well, since the Nash weight of (k,b,c) = (1,56,1) is low (if a triple (k,b,c) does not satisfy all four conditions, then the Nash weight of this (k,b,c) triple is 0).
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Old 2020-02-17, 16:35   #6
R. Gerbicz
 
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Quote:
Originally Posted by Batalov View Post
Well known that finitely few.

Sit down with a piece of paper and think what values can n take for 56^n+1 to be prime?
Only powers of 2. b^2^m+1 are called generalized Fermat numbers. Read Mathworld for a starter.
There is an exception, for n=0 it is a prime.
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