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.