20171217, 18:15  #1 
Nov 2016
29 Posts 
Final Lucas Lehmer residuals and Pythagorean triples
Bas Jansen's phd paper has a table of Lehmer symbols:
https://www.math.leidenuniv.nl/scripties/PhDJansen.pdf (4,q) [5(+),7(),13(+),17(),19(),31(+),61(+),89(),107(),127(+),521()] (10,q) [5(),7(),13(),17(+),19(+),31(+),61(+),89(+),107(+),127(+),521(+)] The rows represent the sign table of the Lehmer symbols. The first row is formed from the residuals of a LL sequence with a starting value of 4. The second row is a starting value of 10. I've clipped and attached a proper definition of the Lehmer symbol below (Lehmer symbols are the major subject of the phd). Pythagorean triples and Lucas Lehmer residuals Using the Mersenne primes as an example: Calculate the second to last residual from a LL sequence with a starting value of 4: [8,111,128,130559,523263,65536,2147483648,618970019642654953077473279,162259276829213345377179500806143,18446744073709551616,...] In binary: [0b1000,0b1101111,0b10000000,0b11111110111111111,0b1111111101111111111,0b10000000000000000,...] Taking those residuals with a () Lehmer symbol: Exponents: [7,17,19,89,107,521,607,1279,2281,3217,4423,9689,11213...] Residuals(s): [111,130559,523263,618970019642654953077473279,162259276829213345377179500806143,...] I noticed these residuals seem to be one and two less than b and c in a Pythagorean triple, as in: a1 = sqrt(c1^2b1^2) = sqrt(113^2112^2) = 15.0 b1 = s+1 c1 = s+2 Triples: [15,112,113],[511,130560,130561],[1023,523264,523265],[35184372088831,618970019642654953077473280,618970019642654953077473281]... In the LL test, the second to last residual is squared, then 2 is subtracted. This number is then reduced modulo the Mersenne number and the modulo returns 0 if the test finds a prime, as in: 12319 mod 2^71 = 0 In the case of negative Lehmer symbols, this squaring also generates the b value of another Pythagorean triple: a2 = 2*a1*b1 = 2*15*112 = 3360 b2 = (b1a1)*(b1+a1) = (11215)*(112+15) = s^22 = 111^22 = 12319 c2 = c1^2 = 113^2 = 12769 sqrt(12769^2  12319^2) = 3360.0 Triples: [3360,12319,12769],[133432320,17045652479,17046174721],[1070598144,273804167167,273806260225]... I've tested up to 2^112131 and all those listed above have a2^2 + b2^2 = c2^2 in the case of a starting value of 4. (+) Lehmer symbols do not produce integer value triples via the above method. However, it is possible to manually generate these triples for composite and (+) Mersenne numbers, as in: a1=63, b1=1984, c1=1985 a2=249984, b2=3932287, c2=3940225 3932287 mod 2^111 = 0 I just thought I'd pass this on, I haven't found it mentioned elsewhere. Please let me know if anything doesn't check out. Last fiddled with by a nicol on 20171217 at 19:11 
20171218, 00:14  #2 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20171218, 13:25  #3  
Nov 2016
1D_{16} Posts 
Quote:
I think these triples exist for all Mersenne numbers, I just thought it might be worth cataloguing the coincidence between them and the LL test & Lehmer symbols. I think there might be a tree to follow here? I think that would be worth looking at? https://en.wikipedia.org/wiki/Tree_o...gorean_triples 

20171218, 13:29  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20171218, 16:06  #5  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9333_{10} Posts 
Quote:
It is 2^{p}  1  2^{(p+1)/2}, or in other words 2X^{2}  2X  1 with X=2^{(p1)/2}. Of course there is a Pythagorean triple {a, 2X^{2}  2X, 2X^{2}  2X + 1}. Now, find a. It is 2X1. This is similar at looking at the definition of primes, writing a dozen of them, ...and then 'noticing' that they just happen to be numbers that are only divisible by themselves and 1. Or 'noticing' that they are all odd if they are > 2. Everything else (' ...it is possible to manually generate these triples for composite and (+) Mersenne numbers') is handwaving. 

20171218, 18:39  #6  
Nov 2016
29 Posts 
Quote:
There is also the fact that the Lucas Lehmer sequences themselves are Heronian consecutive integer triangles  14,194,37634,1416317954,2005956546822746114 are the b values for: [13,14,15],[193,194,195],[37633,37634,37635] Maybe there's also a trivial answer to why modding a Mersenne prime against one of the values of a Heronian triangle will eventually output another integer area triangle (some of the time)? 

20171218, 19:05  #7 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2^{2}·3·883 Posts 

20171218, 19:31  #8 
Nov 2016
29 Posts 

20171218, 22:12  #9 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
the URL you started with has some for the irregular start values.
Last fiddled with by science_man_88 on 20171218 at 22:13 
20171218, 23:58  #10 
Nov 2016
29 Posts 
Why, mathematically speaking, do these signs vary depending on the start value? Seems like there's a very large dense class theoretical answer to that. Maybe someone could help simplify the answer with some kind of visual analogy?

20171219, 00:12  #11  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}·17·61 Posts 
Quote:
Quote:
If you want the same for s_{0} = 10, set InitialLLValue=10 in prime.txt. If you want the same for s_{0} = 2/3, set InitialLLValue=23 in prime.txt Just go and read there  http://mersenneforum.org/showpost.ph...9&postcount=30 ____ * Google for "Now that you've found it, it's gone / Now that you feel it, you don't". No, there is no hidden meaning in this footnote. Most of the time, the point of that poem is true  for everyone. 

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