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Old 2015-10-13, 08:27   #1
kalikidoom
 
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Default Official 'exchange of inanities' thread [Was: mm127 is prime, cuz I say so]

...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?

Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
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Old 2015-10-13, 13:22   #2
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Quote:
Originally Posted by kalikidoom View Post
...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?

Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
That is false. For n=4, all are composite. Also for n=11, all are composite.
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Old 2015-10-13, 13:25   #3
R.D. Silverman
 
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Quote:
Originally Posted by kalikidoom View Post
...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?
I see a sequence of numbers. I see no mathematical argument at all that suggests
every number in the sequence is prime. You have a strange notion about what
constitutes a proof.

Quote:
Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
"It always works". Is this your proof? A simple assertion?

I suggest that you think about what you wrote.

For example, please prove to us that 2^(2^127-1) - 1 is prime.

Last fiddled with by R.D. Silverman on 2015-10-13 at 13:28
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Old 2015-10-13, 13:26   #4
retina
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Quote:
Originally Posted by kalikidoom View Post
<blah>... it always works
For various values of "works".
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Old 2015-10-13, 13:32   #5
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Quote:
Originally Posted by kalikidoom View Post
...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?

Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
1) n must be prime for 2^n-1 to be prime
2) by that same logic m=2^n-1 must be prime for 2^m-1= 2^(2^n-1)-1 to have a chance at being prime.
3) as LaurV pointed out not all n that are prime will allow 2^n-1 to be prime.
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Old 2015-10-13, 13:59   #6
R.D. Silverman
 
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Quote:
Originally Posted by LaurV View Post
That is false. For n=4, all are composite. Also for n=11, all are composite.
I expect that the OP intended for n = 2.

However, I'd still like an answer to my question as to why the OP thinks that presenting a
sequence of numbers is in any way a "proof". There were no mathematical statements
asserting that some (set of) condition(s) is true, nor were there any logical statements.
It was just a list of numbers. How can this be a proof?
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Old 2015-10-13, 14:03   #7
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Well, the original poster said:
Quote:
Originally Posted by kalikidoom View Post
...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?
He didn't say that he has a proof.
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Old 2015-10-13, 14:14   #8
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Quote:
Originally Posted by alpertron View Post
Well, the original poster said:

He didn't say that he has a proof.
He suggested that a sequence of numbers constituted a proof.
It is hard to correct someone's misconceptions without knowing how and
why they believe what they believe. A simple statement that what was
posted is not a proof says very little./// We need to know why the OP thought
that simply presenting a sequence of numbers might be a proof.

Last fiddled with by R.D. Silverman on 2015-10-13 at 14:14 Reason: typo
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Old 2015-10-13, 14:24   #9
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Quote:
Originally Posted by R.D. Silverman View Post
He suggested that a sequence of numbers constituted a proof.
It is hard to correct someone's misconceptions without knowing how and
why they believe what they believe. A simple statement that what was
posted is not a proof says very little./// We need to know why the OP thought
that simply presenting a sequence of numbers might be a proof.
one thought I had is that they might think that because you can infinitely add to the representation that it represents and infinite amount of numbers and might intersect the primes infinitely often.
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Old 2015-10-13, 14:25   #10
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Quote:
Originally Posted by science_man_88 View Post
one thought I had is that they might think that because you can infinitely add to the representation that it represents and infinite amount of numbers and might intersect the primes infinitely often.
Gibberish from someone who hasn't yet passed first year algebra.

Last fiddled with by R.D. Silverman on 2015-10-13 at 14:26
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Old 2015-10-13, 14:29   #11
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Quote:
Originally Posted by R.D. Silverman View Post
Gibberish from someone who hasn't yet passed first year algebra.
I didn't say it was mathematically sound, I was trying to think like the OP might be thinking. Yet assumes I plan on going back to school again.
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