20210828, 15:20  #1 
"ม้าไฟ"
May 2018
212_{16} Posts 
"I am experimenting with heuristic sieves."
2^{1277}  1 = 1 + 2⋅3⋅5⋅23⋅59⋅89⋅233⋅397⋅683⋅1103⋅1277⋅2089⋅2113⋅18503⋅64439⋅181193⋅3033169⋅107367629⋅...
https://en.wikipedia.org/wiki/Fermat%27s_little_theorem Last fiddled with by Dobri on 20210828 at 15:36 
20210828, 17:03  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}×3×5^{2}×17 Posts 
Quote:
What does it have to do with anything? 

20210828, 17:50  #3 
"ม้าไฟ"
May 2018
2·5·53 Posts 

20210828, 18:07  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
23730_{8} Posts 
Quote:
Ever heard of x^{2}1 = ... ? (write it out, don't be shy) x^{11}1 = ... ? (write it out) There is so much "interesting" in the world to anyone who skipped all classes in the 7th grade. 

20210828, 18:30  #5  
"ม้าไฟ"
May 2018
2×5×53 Posts 
Quote:
One should not forget to be polite and respectful to the ones who attempt to communicate instead of posting uncivilized forcible content of provocative and insulting nature as only bullies do from the assumed position of power. At least, I remember from kindergarten the basics of being kind to people. 

20210828, 19:13  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}×3×5^{2}×17 Posts 
You, actually, appear to be assuming the worst in others.
I almost always apply Hanlon's_razor. My remark had nothing to do with you but with your argument. Do you disagree with an illustrative argument that if a person cut all classes in school in the past, they will later in life find a lot of things interesting and even stunning? "A kilogram of led weighs the same as a kilogram of feathers! Who could have thought?!" You are arguing not with the presented argument, but with the "assumed" fact that it was "about you". It was not. Whether you immediately project yourself in every sentence is entirely a matter of your choice. Your assumptions. 
20210828, 20:09  #7 
"ม้าไฟ"
May 2018
2×5×53 Posts 
Same here. Only the first line of my previous response was concerned with your comment concerning the homogeneous version of the geometric sum formula.
The rest of said response was an illustrative argument. As for razors, I do not use such tools. Sometimes, one could inadvertently 'cut' oneself while attempting to apply razors to others. 
20210828, 20:55  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·3·5^{2}·17 Posts 
I can only add another illustrative vignette 
Interestingly (c), 2^{1278}1 is fully factored. But how can that be without the help of factoring 2^{1278}11, which is not factored at all beyond the trivial factor of 2? It is one of the two: 1. Wizardry! or 2. Factoring N1 actually helps nothing to factor N; it is a red herring. 
20210828, 21:17  #9  
"ม้าไฟ"
May 2018
2×5×53 Posts 
Quote:
Every heuristic initially has an element of wizardry though. 

20210828, 21:50  #10  
Feb 2017
Nowhere
18F3_{16} Posts 
Quote:
I see no indication that your "heuristic sieve" is informed by anything, or that it is anything other than phantasmagorical. You have provided no motivation for using factors of M_{p}  1 in a sieve for trying to factor M_{p}, and have given absolutely no indication of what set, if any, you might want to use them to sieve. You are certainly not informed on the subject by what is available on this Forum. The question of whether factoring M_{p}  1 would help in factoring M_{p} was discussed on this very forum about a year ago, here. It specifically mentions the exponent 1277. And, as of this posting, the thread title is still visible on the Mersenne Forum home page. The short answer to the question is "No." I conclude that you have either not bothered to avail yourself of the most basic facts concerning factors of M_{p}  or are deliberately disregarding them, in which case you are a troll. 

20210828, 23:04  #11  
"Tucker Kao"
Jan 2020
Head Base M168202123
1560_{8} Posts 
Quote:
x^{3}1 = (x  1)(x^{2} + x + 1) x^{3}+1 = (x + 1)(x^{2}  x + 1) x^{6}1 = (x^{3}1)(x^{3}+1) x^{11}1 = didn't get any Google Search results, but since there are answers for x^{5}1 and x^{5}+1, the same patterns should be observable. x^{5}1 = (x  1)(x^{4} + x^{3} + x^{2} + x + 1) x^{5}+1 = (x + 1)(x^{4}  x^{3} + x^{2}  x + 1) x^{2p}+1 = Maybe a prime x^{2p}1 = Always composite x^{2p±1}+1 = Always composite x^{2p±1}1 = Always composite Last fiddled with by tuckerkao on 20210829 at 00:03 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Sieving success as a heuristic for odds of a prime  diep  Probability & Probabilistic Number Theory  0  20201121 17:14 
rogue's sieves  rogue  And now for something completely different  4  20170731 19:36 
Experimenting with ksieve  Cruelty  Riesel Prime Search  18  20060625 03:44 