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Topic: My next math challenge! / close semiprimes
Replies: 3   Last Post: Jul 16, 2014 4:57 AM

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James Waldby

Posts: 358
Registered: 1/27/11
Re: My next math challenge! / close semiprimes
Posted: Jul 15, 2014 11:56 PM
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On Wed, 16 Jul 2014 03:20:11 +0000, James Waldby wrote:
> On Tue, 15 Jul 2014 11:58:54 -0700, djoyce099 wrote:
...
>> I will have to check this out because I believe four discrete same length
>> primes that create (two separate semi-primes) can only have a difference
>> between semi-primes > sqrt(largest prime). Whether or not the 4 primes are twins. Your claim is that two semi-primes that are 616 digits in length
>> can have a difference of only 36936. Are your 4 primes discrete?

>
> There are 4 different primes in each example I gave. Here's a small
> example: 3253*3257 - 3251*3259 = 10595021 - 10595009 = 12 < 57 < sqrt(3259)
> Here are some 20-digit examples which also have a difference of 12 between
> the semiprimes. Each example involves four different primes, from two
> separate twin-prime pairs. This is output from the code shown below.
>
> Len 20. 3200053063*3200053067 = 10240339618815894221 3200053061*3200053069 = 10240339618815894209 d=12
> Len 20. 3200068513*3200068517 = 10240438500694305221 3200068511*3200068519 = 10240438500694305209 d=12
> Len 20. 3200075143*3200075147 = 10240480933646771021 3200075141*3200075149 = 10240480933646771009 d=12
> Len 20. 3200090473*3200090477 = 10240579048185725621 3200090471*3200090479 = 10240579048185725609 d=12

[snip code]

BTW, note that the difference of 12 is minimal. Each of these product
differences involves a "prime quadruplet" using the term as defined in
the wikipedia article <http://en.wikipedia.org/wiki/Prime_quadruplet>,
ie a set of four primes of form {p, p+2, p+6, p+8}. It isn't known whether
infinitely many prime quadruplets exist. The largest known prime
quadruplet has 3503 digits, which would yield two semiprimes of 7006
digits each, with a difference of 12 between the semiprimes because
(p+6)*(p+2) - p*(p+8) = 12.

--
jiw




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