A question about primes of a particular form

[enzocreti]

here the list up to 10^6 of primes not dividing any pg(k):
1321 3191 3541 23311 49297 87211 131071 476401 823481 870031

[enzocreti]

the prime 23311 divides 2^(3511-1)-1 where 3511 is a Wieferich prime

[enzocreti]

49297=13^4+12^4

[science_man_88]

Originally Posted by enzocreti

49297=13^4+12^4

https://en.m.wikipedia.org/wiki/Sum_...quares_theorem

[enzocreti]

the pg primes 31,73,157 are factors of 2^(3511-1)-1 and 2^(1093-1)-1, where 3511 and 1093 are the known Wieferich primes

[Dr Sardonicus]

Originally Posted by enzocreti

here the list up to 10^6 of primes not dividing any pg(k):
1321 3191 3541 23311 49297 87211 131071 476401 823481 870031

Either your code ain't writ right, or my code ain't writ right, or you made an oopsadaisy with transcription. The prime 459691 seems to fill the bill, but is not on your list.

Also -- if you define pg(1) = 10, then the prime 2 divides pg(1) but not p(k) for any k > 1. If you require k > 1, then the prime 2 goes to the head of the list.

[Batalov]

>5 messages every hour?
This is not even misc math. This is clearly blogorrhea

[science_man_88]

Originally Posted by Batalov

>5 messages every hour?
This is not even misc math. This is clearly blogorrhea

I agree, especially small postings that could be merged.

[enzocreti]

With Pari I tried to find primes p of the form:


(7^q+1)/(7^(q-n^2)+1)=p where q and n are positive integers.
Up to q=10000 I found only four solutions for q=17,24,38,148 and n=4,4,6,12.
The first question is: do you believe that the number of these primes is not infinite?
Then do you believe that (q-n^2) is always of the form 2^j, where j is an integer equal or grater than zero?
And generalizing:
what about primes p of the form:


(s^q+1)/(s^(q-n^2)+1)=p where s is any prime?

[axn]

Originally Posted by enzocreti

Then do you believe that (q-n^2) is always of the form 2^j, where j is an integer equal or grater than zero?

Almost. Basically you're looking for numbers of the form a^n+1 with only one algebraic factorization. This means n will be of the form p^i*2^j, p is prime, i>=1, j >=0. If n has more than one odd prime factor, then it will have two or more algebraic factors, but you are only dividing out one algebraic factor, so it cannot be prime.

Next, q-n^2 must divide q. In fact, q-n^2 must be equal to q/p (p being the sole odd prime factor of q).
Now, n^2 = q-q/p = q*(p-1)/p = p^(i-1)*2^j*(p-1). For this to be a square, i-1 should be even and either (p-1) should be a power of two and 2^j*(p-1) should be a square or (p-1) should be a square and j should be even.

That will get you eligible candidates (without trivial algebraic factors). I think if you search enough candidates for enough bases, you'll see something where q-n^2 is not a power of two.

EDIT:- Try searching primes of the form (x^125+1)/(x^25+1) . Here 25 = 125-10^2. It is prime for x = 2081, 2753, 3253, 5591, 6073 (only prime x were considered. composite x can also yield primes of this form.)

[enzocreti]

believe it or not this number is certified prime:

(2^4871-479)/(4871+2002) with 1463 digits!!!
This is wonderful Amazing!!!