Reinventing the bicycle about Mersenne prime
 

My discovery about Mersenne prime
I find there are a lot of numbers similar to Mersenne prime and Fermat number.
(a+1)^n-a^n，(a+b)^(2^n)+a^(2^n)
Looking at [url]http://weibo.com/ttarticle/p/show?id=2309404101129943270060[/url]

[QUOTE=353085177;457639]My discovery about Mersenne prime
I find there are a lot of numbers similar to Mersenne prime and Fermat number.
(a+1)^n-a^n，(a+b)^(2^n)+a^(2^n)
Looking at [url]http://weibo.com/ttarticle/p/show?id=2309404101129943270060[/url][/QUOTE]

(a+1)^2-(a^2) describes all odd numbers as it's 2*a+1 turns out the form the link has is (a+1)^2+a^2. a square can only be 0,1, or 4 mod ( remainder on division by) 8. so without caring for the order they show up we can have 0+0,0+1,0+4,1+1,1+4,4+4 = 0,1,4,2,5,0 remainders as sums of squares in general.

[QUOTE=science_man_88;457652](a+1)^2-(a^2) describes all odd numbers as it's 2*a+1 turns out the form the link has is (a+1)^2+a^2. a square can only be 0,1, or 4 mod ( remainder on division by) 8. so without caring for the order they show up we can have 0+0,0+1,0+4,1+1,1+4,4+4 = 0,1,4,2,5,0 remainders as sums of squares in general.[/QUOTE]

You're right.

anything else to think about ? oh yes only if a and b are opposite parity will (a+b)^(2^n)+a^(2^n) be odd. edit: okay not technically true if a is even then the second part is even and you need b to be odd for there to be both an odd and even part to the sum if a is odd then b being odd is allowed as that makes a+b even leading to an even +odd = odd result.

[QUOTE=science_man_88;457659]anything else to think about ? oh yes only if a and b are opposite parity will (a+b)^(2^n)+a^(2^n) be odd. edit: okay not technically true if a is even then the second part is even and you need b to be odd for there to be both an odd and even part to the sum if a is odd then b being odd is allowed as that makes a+b even leading to an even +odd = odd result.[/QUOTE]

if a is even and b is odd = odd result
if a is odd and b is even = even result

[QUOTE=353085177;457662]if a is even and b is odd = odd result
if a is odd and b is even = even result[/QUOTE]

yes well the point is unless the powers are opposite parities the sum is even ( I realise my mistake as talking about b when I should have said a+b and a have to be opposite parities).

[QUOTE=science_man_88;457666]yes well the point is unless the powers are opposite parities the sum is even ( I realise my mistake as talking about b when I should have said a+b and a have to be opposite parities).[/QUOTE]

oh

[QUOTE=353085177;457670]oh[/QUOTE]

what else have you discovered about the mersenne numbers ?

like have you read about their factors for prime exponents ?
have you figured out that you can make the LL test similar to the trial factoring done ?

etc.

I think you'll find that factors of fermat numbers ae k*2^(n+2)+1 if I remember reading correct ( it's also how the small factor I found of F10 using PARI/GP works out. I'll be back later tonight maybe.

turns out I'm doing so much work related math as well that I forgot I don't work today.

[QUOTE=science_man_88;457682]turns out I'm doing so much work related math as well that I forgot I don't work today.[/QUOTE]

I know little about factor

[QUOTE=353085177;457692]I know little about factor[/QUOTE]

[url]http://mersenneforum.org/showthread.php?t=17126[/url] may help. though you list the form of factor much can be said about the k for specific p.

Just a note to the OP, to reinvent means:

"change (something) so much that it appears to be entirely new."

Admittedly, I should tell myself this more often as well. It's hard to change something so it looks new without first knowing the old.

[QUOTE=science_man_88;457694][url]http://mersenneforum.org/showthread.php?t=17126[/url] may help. though you list the form of factor much can be said about the k for specific p.[/QUOTE]

Finding primes of (a+1)^n-a^n, a>1,harder than 2^n-1,we need to find a new way.

[QUOTE=353085177;457736]Finding primes of (a+1)^n-a^n, a>1,harder than 2^n-1,we need to find a new way.[/QUOTE]

2^n-1 is a form of the former ( at least without the restriction) so really what's needed is a generalized way. technically by fermat's little theorem [TEX]a^(p-1)\equiv 1 mod p[/TEX] when a and p are coprime. so (a+1)^n-a^n a>1 allows us to use this to say it's congruent to [TEX](a+1)^{(n%(p-1))}-(a^{(n%(p-1))}) [/TEX]where n%(p-1) is the remainder of n when divided by p-1 whenever both a and a+1 are coprime to the prime we are testing it for factors with.

[QUOTE=353085177;457736]Finding primes of (a+1)^n-a^n, a>1,harder than 2^n-1,we need to find a new way.[/QUOTE]
The collective "we" already have. (Not really a new way, but then again what's new for someone is old for someone else.)

Did you see these - [URL]http://www.primenumbers.net/prptop/searchform.php?form=b%5En-a%5En&action=Search[/URL] ?

[QUOTE=Batalov;457743]The collective "we" already have. (Not really a new way, but then again what's new for someone is old for someone else.)

Did you see these - [URL]http://www.primenumbers.net/prptop/searchform.php?form=b%5En-a%5En&action=Search[/URL] ?[/QUOTE]

But he did not realize the real Generalized Mersenne and Generalized Fermat,he summed up the wrong.[URL="http://www.primenumbers.net/Henri/us/MersFermus.htm"]http://www.primenumbers.net/Henri/us/MersFermus.htm[/URL]

[QUOTE=science_man_88;457738]2^n-1 is a form of the former ( at least without the restriction) so really what's needed is a generalized way. technically by fermat's little theorem [TEX]a^(p-1)\equiv 1 mod p[/TEX] when a and p are coprime. so (a+1)^n-a^n a>1 allows us to use this to say it's congruent to [TEX](a+1)^{(n%(p-1))}-(a^{(n%(p-1))}) [/TEX]where n%(p-1) is the remainder of n when divided by p-1 whenever both a and a+1 are coprime to the prime we are testing it for factors with.[/QUOTE]

I'm sorry I can't understand.

[QUOTE=353085177;457747]I'm sorry I can't understand.[/QUOTE]

[url]https://en.wikipedia.org/wiki/Modular_exponentiation[/url]

[QUOTE=science_man_88;457758][url]https://en.wikipedia.org/wiki/Modular_exponentiation[/url][/QUOTE]

n%(p-1),n=?,p=?Can you give me an example?

[QUOTE=353085177;457762]n%(p-1),n=?,p=?Can you give me an example?[/QUOTE]

basically fermat's little theorem states that p being a prime, that for any a, coprime to p, [TEX]a^p \equiv a \pmod p[/TEX] this directly translates as [TEX]a^{(p-1)}\equiv 1 \pmod p[/TEX] so for example with p =3 ; a=4; and n =101 we can say using the fact that 1^n for any n is 1 that since 101 = 2*50+1 and 2 is p-1. that 4^101-1 is divisible by 3 coming indirectly from the fact that 4^1-1 is divisible by 3. 4^101 = (4^2)^50 * 4^1 -1 ;4^2 is 1 mod 3 and 1^n is 1 for any n so the whole first part reduces to 1*4^1-1 = 4^1-1 so it will have the same modular remainder as 4^1-1 when trying mod 3. edit: in fact we if the base is higher than the prime in question we can mod the bas as well so it reduces further to 1^1-1 = 1-1 =0. so the remainder mod 3 is 0.

[QUOTE=science_man_88;457766]basically fermat's little theorem states that p being a prime, that for any a, coprime to p, [TEX]a^p \equiv a \pmod p[/TEX] this directly translates as [TEX]a^{(p-1)}\equiv 1 \pmod p[/TEX] so for example with p =3 ; a=4; and n =101 we can say using the fact that 1^n for any n is 1 that since 101 = 2*50+1 and 2 is p-1. that 4^101-1 is divisible by 3 coming indirectly from the fact that 4^1-1 is divisible by 3. 4^101 = (4^2)^50 * 4^1 -1 ;4^2 is 1 mod 3 and 1^n is 1 for any n so the whole first part reduces to 1*4^1-1 = 4^1-1 so it will have the same modular remainder as 4^1-1 when trying mod 3. edit: in fact we if the base is higher than the prime in question we can mod the bas as well so it reduces further to 1^1-1 = 1-1 =0. so the remainder mod 3 is 0.[/QUOTE]

4^101 = (4^2)^50 * 4^1 -1? not 4^101 = (4^2)^50 * 4^1?And what do we know at last?

[QUOTE=353085177;457770]4^101 = (4^2)^50 * 4^1 -1? not 4^101 = (4^2)^50 * 4^1?And what do we know at last?[/QUOTE]

sorry typo 4^101-1 equals (4^2)^50*(4^1) -1 which reduces to 1^50*(4^1)-1 = 1*(4^1)-1 which is just 4^1-1 = 3 is divisible by 3.

[QUOTE=science_man_88;457771]sorry typo 4^101-1 equals (4^2)^50*(4^1) -1 which reduces to 1^50*(4^1)-1 = 1*(4^1)-1 which is just 4^1-1 = 3 is divisible by 3.[/QUOTE]

Is used to determine whether the 3 factor is not 5^101-4^101,?I still do not understand.I'm not a math major.

[QUOTE=353085177;457773]Is used to determine whether the 3 factor is not 5^101-4^101,?I still do not understand.I'm not a math major.[/QUOTE]

neither am I I've just been on this forum for closing in on 8 years. well yes you can because you can prove something about each part: [TEX]5^{101} -4^{101} \equiv 2^{(101%2)}-1^{(101%2)} \pmod 3 \equiv 2^1-1^1 \equiv 1 \pmod 3[/TEX]

[QUOTE=science_man_88;457776]neither am I I've just been on this forum for closing in on 8 years. well yes you can because you can prove something about each part: [TEX]5^{101} -4^{101} \equiv 2^{(101%2)}-1^{(101%2)} \pmod 3 \equiv 2^1-1^1 \equiv 1 \pmod 3[/TEX][/QUOTE]

But any factor p of (a+1)^n-a^n must be of the form 2kn+1,n%(p-1)=?

[QUOTE=353085177;457785]But any factor p of (a+1)^n-a^n must be of the form 2kn+1,n%(p-1)=?[/QUOTE]

the remainder of n or divisions by p-1 is what n%(p-1) is. Right, ... But, you can use the same style of argument. if a is greater than n then you can take the remainder by this value p so you could reduce the base for any divisor talked about in these cases.

[QUOTE=science_man_88;457787]the remainder of n or divisions by p-1 is what n%(p-1) is. Right, ... But, you can use the same style of argument. if a is greater than n then you can take the remainder by this value p so you could reduce the base for any divisor talked about in these cases.[/QUOTE]

P is the form 2kn+1, n% (p-1) =0, the final results are = 0 (mod p)……

[QUOTE=353085177;457788]P is any prime number, n% (p-1) =0, the final results are = 0 (mod p)……[/QUOTE]

okay, so that extension doesn't apply if n<p which if p=2*k*n+1 is always true. but you can still reduce the bases if they are larger than p.

[QUOTE=science_man_88;457789]okay, so that extension doesn't apply if n<p which if p=2*k*n+1 is always true. but you can still reduce the bases if they are larger than p.[/QUOTE]

But if p=2*k*n+1,n must be less than p

[QUOTE=353085177;457791]But if p=2*k*n+1,n must be less than p[/QUOTE]

okay but in the case of 505^101-504^101 you can show this has remainder 183 on division by 203 because that's what 99^101-98^101 is so the reducing the base part can still apply even for these numbers. I guess I don't see what's new in your page.

It's time to move to Misc.Math.

[QUOTE=Batalov;457805]It's time to move to Misc.Math.[/QUOTE]

agreed.

the whole thing divides by 2*a+1 when n is odd ( (a+1)^n+a^n , I mean by whole thing), as a+1 can be restated in the remainder math as -a, giving (-a)^n+(a^n) with this modulus ( as it's called). so (5^101+4^101) divides by 9. the difference side of things (a+b)^n-a^n is always divisible by (a+b)-a=b so only if b is 1 is there not a trivial factor.

[QUOTE=353085177;457639]My discovery about Mersenne prime
I find there are a lot of numbers similar to Mersenne prime and Fermat number.
(a+1)^n-a^n，(a+b)^(2^n)+a^(2^n)
Looking at [url]http://weibo.com/ttarticle/p/show?id=2309404101129943270060[/url][/QUOTE]

There are already plenty of generations to Mersenne numbers only using polynomials P(2^x).

Let n be any odd number, and the sequence 2^n-1 = 1, 7, 31, 127, 511, 2047, 8191, 32767, 131071, 524287,......

1) switch to (2^n+1)/3 for all odd n is another generalization:

1, 3, 11, 43, 171, 683, 2731, 10923, 43691, 174763,......

2) 2^n+-2^((n+1)/2)+1 defines a similar sequence depending on n = (1, 7) or (3, 5) (mod 8):

1, 13, 41, 113, 481, 2113, 8321, 32513, 130561, 525313,......

3) (2^(d*n)-1)/((2^d-1)*(2^n-1)) is another generalize prime exponent form.

[QUOTE=carpetpool;457852]There are already plenty of generations to Mersenne numbers only using polynomials P(2^x).

Let n be any odd number, and the sequence 2^n-1 = 1, 7, 31, 127, 511, 2047, 8191, 32767, 131071, 524287,......

1) switch to (2^n+1)/3 for all odd n is another generalization:

1, 3, 11, 43, 171, 683, 2731, 10923, 43691, 174763,......

2) 2^n+-2^((n+1)/2)+1 defines a similar sequence depending on n = (1, 7) or (3, 5) (mod 8):

1, 13, 41, 113, 481, 2113, 8321, 32513, 130561, 525313,......

3) (2^(d*n)-1)/((2^d-1)*(2^n-1)) is another generalize prime exponent form.[/QUOTE]

as for the actual primes themselves they also have generalizations in forms:

[url]https://en.wikipedia.org/wiki/Mersenne_prime#Generalizations[/url]