if p is prime, factors of 2^p-1? - Is it possible?

[LaurV]

Originally Posted by ONeil

A composite number produced from a prime number using 2^p-1 does contain prime factors!

What is that?

How do you "produce a composite number from a prime number"? Do you multiply it with 7? Or subtract 53? Can you prove that, by doing so, you don't end with another number which is still prime?

And you want to do that by "using 2^p-1". How? You take 2^p-1 and hit your prime number in the head with it until it becomes composite?

And what the hack is "contains prime factors" having to do with the way you "produced" it? All composite numbers contain prime factors, regardless of how you "produced" them. All prime numbers "contain prime factors" too (themselves).

Man, how old are you? You may be like 12 years old or so, learning the basics right now, and not having the proper way to express yourself in English yet, and in that case, you may be a genius and become great in the future. At that age I didn't have any idea about all this stuff. If that's the case, my apologies.

But if you are older than 18, stop the funking trolling and go learn the lingo, if you want to talk math.

P.S. I never pretended to be a decent person.

[Viliam Furik]

I would love a way to give a "like" or some kind of positive evaluation of my opinion to the post, without having to write another one, saying I like it. Hmmm, Xyzzy ?

[Dr Sardonicus]

Originally Posted by Viliam Furik

This works with only k=1, thus if it divides 2^p-1, it is definitely a prime factor.

Actually, it's true for more than just k = 1.

Let p > 2 be prime. The smallest conceivable composite factor of 2^p - 1 is (2p+1)².

So if q = 2*k*p + 1 divides 2^p - 1, and k < 2*p + 2, then q is prime.

Alas, it is more than likely that if k < 2*p + 2 and q = 2*k*p + 1 is prime, that q does not divide 2^p - 1.

Especially if q is congruent to 3 or 5 (mod 8)

[LaurV]

Originally Posted by Dr Sardonicus

So if q = 2*k*p + 1 divides 2^p - 1, and k < 2*p + 2, then q is prime.

Respective k<6*p+1 when p=1 (mod 4)

[Dr Sardonicus]

Originally Posted by LaurV

Respective k<6*p+1 when p=1 (mod 4)

[ONeil]

Originally Posted by LaurV

What is that?

How do you "produce a composite number from a prime number"? Do you multiply it with 7? Or subtract 53? Can you prove that, by doing so, you don't end with another number which is still prime?

And you want to do that by "using 2^p-1". How? You take 2^p-1 and hit your prime number in the head with it until it becomes composite?

And what the hack is "contains prime factors" having to do with the way you "produced" it? All composite numbers contain prime factors, regardless of how you "produced" them. All prime numbers "contain prime factors" too (themselves).

Man, how old are you? You may be like 12 years old or so, learning the basics right now, and not having the proper way to express yourself in English yet, and in that case, you may be a genius and become great in the future. At that age I didn't have any idea about all this stuff. If that's the case, my apologies.

But if you are older than 18, stop the funking trolling and go learn the lingo, if you want to talk math.

P.S. I never pretended to be a decent person.

Your comprehension is weak go back to school and take a math class.

[Batalov]

Originally Posted by ONeil

Your comprehension is weak go back to school and take a math class.

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