is the factorisation of Mp1 an advantage ?
A peaceful and pleasant night for you,
if I know the factorisation or a part of the factorisation of Mp1 do I have any advantages for checking the primality ? (Mp should be a Mersenne number) Or in other words, is the factorisation of p1 helpful ? I know the theorem of Pocklington for proofing primality [url]https://en.wikipedia.org/wiki/Pocklington_primality_test[/url] Thanks in advance if you spend me some lines :cmd: :truck: :uncwilly: Bernhard 
[QUOTE=bhelmes;553693]
if I know the factorisation or a part of the factorisation of Mp1 do I have any advantages for checking the primality ? [/QUOTE] Probably there is no advantage for that, but any odd factor of Mp1 could give a nontrival factor of another Mersenne number (with prime index), since rMp1=2*(2^(p1)1). 
[QUOTE=bhelmes;553693]Or in other words, is the factorisation of Mp1 helpful ?[/QUOTE]
N=Mp are a beautiful example of being proven with >>33.33% N[B]+1[/B] factorization (indeed, 100%). Factorization of N1 is needless, when you have a 100% N[B]+1[/B] factorization. 
Agree with Batalov; for proving primality of M_p, since the full factorization of M_p + 1 is trivial, we do not gain anything from the factorization of M_p  1.
Of course, it may be fun to find the factorization anyway; [URL="http://factordb.com/index.php?query=2%5En3&use=n&n=2&VP=on&EV=on&OD=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]here is a factordb query for tiny examples[/URL]. /JeppeSN 
[QUOTE=JeppeSN;553877]Of course, it may be fun to find the factorization anyway; [URL="http://factordb.com/index.php?query=2%5En3&use=n&n=2&VP=on&EV=on&OD=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]here is a factordb query for tiny examples[/URL].[/QUOTE]
You linked to Mp2. 
[QUOTE=axn;553881]You linked to Mp2.[/QUOTE]
Oops, that is right. It should have been [URL="http://factordb.com/index.php?query=2%5En2&use=n&n=2&VP=on&EV=on&OD=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]2^n2 for n prime[/URL], or 2*(2^(n1)  1). /JeppeSN 
A peaceful and pleasant day in spite of Covid19
[QUOTE=JeppeSN;554050][URL="http://factordb.com/index.php?query=2%5En2&use=n&n=2&VP=on&EV=on&OD=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]2^n2 for n prime[/URL][/QUOTE] I noticed that the factors of Mp1 have the following proporty: r odd and r  p1 then the factors f of Mp1 with f > p are f=1 mod r I was a little bit astonished about it. I have no explication for it. Perhaps someone could explain it to me and other members, interested in number theory. :hello: :cmd: :uncwilly: @Batalov: The factoring for Mp+1 is easy, sufficient for a proof and very nice, but perhaps the factoring of Mp1 gives some more details. 
See the exercise [URL="https://www.mersenneforum.org/showpost.php?p=477146&postcount=3"]here[/URL]

[QUOTE=bhelmes;572322]I noticed that the factors of Mp1 have the following proporty:
r odd and r  p1 then the factors f of Mp1 with f > p are f=1 mod r[/QUOTE] Not quite. M12771 has a factor of 2089, among others. 12771 is divisible by 11. 2089 is 10 (mod 11), not 1. 
\(n=M_{1277}1 = 2^{1277}11 = 2^{1277}2 = 2*(2^{1276}1)\). That's a string of 1276 of 1, followed by a zero. 1276 is divisible by 2, 4, 11, 29. Therefore (by grouping the ones), n should be divisible by all factors of respective mersenne, i.e. 2 (from the last 0), 3 (from M2), 5 (from M4), 23, 89 (from M11), 233, 1103, 2089 (from M29). You can find plenty of combinations which are not true, actually, almost all where you "cross" them. In fact, you can say that the residue is 1 only when you don't cross them (i.e. 23 and 89 are 1 (mod 11), or 233, 1103 and 2089 are 1 (mod 29)), but that is a known property of mersenne factors :smile:

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