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




Re: My next math challenge! / close semiprimes
Posted:
Jul 15, 2014 11:56 PM


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 semiprimes) can only have a difference >> between semiprimes > sqrt(largest prime). Whether or not the 4 primes are twins. Your claim is that two semiprimes 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 20digit examples which also have a difference of 12 between > the semiprimes. Each example involves four different primes, from two > separate twinprime 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



