Are there eligible primes Pn such that n < 1.5 x 10^8

George M · 2018-01-04, 08:14

I asked myself if there were any primes $p_n$ and $p_{n + 1}$ (and $p_{n + 2}$ and so on) such that
$2^{p_s} \equiv 1 \pmod m ; s = n, n+1, n+2,\ldots$ for which $m > 2^{64}$ and no other divisors were less than 2^64.

The ultimate goal here was to find an adjacent pair of primes $p_n$ and $p_{n + 1}$ where if $\log_2 k_s = p_s$ then $k_s - 1$ was prime. I found it unlikely for there to exist a solution if we had to find a triplet of primes instead, and not just a pair.

However the only example, and possibly the only triplet of primes, that I have found is if $n = 1.5 \times 10^8 + 0, 1, 2$ . Do there exist these kinds of primes such that $n < 1.5 \times 10^8$ ??

I call these primes “potential primes” since they have the potential to generate a mersenne prime, for the mersenne number they seem to generate has no divisors less than $2^{64}$ .

axn · 2018-01-04, 08:45

You use an undefined term k_s. What is that?

science_man_88 · 2018-01-04, 12:38

Quote:

Originally Posted by George M

I asked myself if there were any primes $p_n$ and $p_{n + 1}$ (and $p_{n + 2}$ and so on) such that
$2^{p_s} \equiv 1 \pmod m ; s = n, n+1, n+2,\ldots$ for which $m > 2^{64}$ and no other divisors were less than 2^64.

The ultimate goal here was to find an adjacent pair of primes $p_n$ and $p_{n + 1}$ where if $\log_2 k_s = p_s$ then $k_s - 1$ was prime. I found it unlikely for there to exist a solution if we had to find a triplet of primes instead, and not just a pair.

However the only example, and possibly the only triplet of primes, that I have found is if $n = 1.5 \times 10^8 + 0, 1, 2$ . Do there exist these kinds of primes such that $n < 1.5 \times 10^8$ ??

I call these primes “potential primes” since they have the potential to generate a mersenne prime, for the mersenne number they seem to generate has no divisors less than $2^{64}$ .

A few things here , m needs to be odd for the mod to work. It needs to be prime if below 2^128. If it needs to work for all s values given it can't exist due to be fact two mersennes with coprime exponents, can't share a factor. Also why is 2^64 so important here. Most of the primenet range was put up to that point of trial factoring a long while ago at last check.

CRGreathouse · 2018-01-04, 15:19

Quote:

Originally Posted by George M

I asked myself if there were any primes $p_n$ and $p_{n + 1}$ (and $p_{n + 2}$ and so on) such that
$2^{p_s} \equiv 1 \pmod m ; s = n, n+1, n+2,\ldots$ for which $m > 2^{64}$ and no other divisors were less than 2^64.

The ultimate goal here was to find an adjacent pair of primes $p_n$ and $p_{n + 1}$ where if $\log_2 k_s = p_s$ then $k_s - 1$ was prime. I found it unlikely for there to exist a solution if we had to find a triplet of primes instead, and not just a pair.

If I understand you, you're looking for distinct primes p and q such that there is some m > 2^64 with 2^p = 2^q = 1 (mod m) and m has no prime divisors below 2^64. Clearly m must divide 2^p - 1 and 2^q - 1 and hence also gcd(2^p - 1, 2^q - 1) = 2^gcd(p, q) - 1 = 2^1 - 1 = 1. As a result m = 1 so m cannot be greater than 2^64.

2018-01-04, 08:14	#1
George M Dec 2017 2·5² Posts	Are there eligible primes Pn such that n < 1.5 x 10^8 I asked myself if there were any primes $p_n$ and $p_{n + 1}$ (and $p_{n + 2}$ and so on) such that $2^{p_s} \equiv 1 \pmod m ; s = n, n+1, n+2,\ldots$ for which $m > 2^{64}$ and no other divisors were less than 2^64. The ultimate goal here was to find an adjacent pair of primes $p_n$ and $p_{n + 1}$ where if $\log_2 k_s = p_s$ then $k_s - 1$ was prime. I found it unlikely for there to exist a solution if we had to find a triplet of primes instead, and not just a pair. However the only example, and possibly the only triplet of primes, that I have found is if $n = 1.5 \times 10^8 + 0, 1, 2$ . Do there exist these kinds of primes such that $n < 1.5 \times 10^8$ ?? I call these primes “potential primes” since they have the potential to generate a mersenne prime, for the mersenne number they seem to generate has no divisors less than $2^{64}$ . Last fiddled with by George M on 2018-01-04 at 08:27

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2018-01-04, 08:45	#2
axn Jun 2003 5464₁₀ Posts	You use an undefined term k_s. What is that?