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Math - Prime number


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#21 bonanova

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Posted 24 November 2007 - 03:39 AM

But I agree ... any prime number can be written as 6n+/-1.
My argument is ... any prime number can be written as 2n+1.
Actually, mine is easier in that you don't need to consider two cases -- 6n+1 and 6n-1.

I have an even easier one: any prime number can be written as n.
Unfortunately I can't complete the proof.

Musings ...

The proof works for any (3p-1); I wonder why the puzzle writer chose 26.
299999 (300000-1) might be discouraging? -1 trivial? ["prime = n" works there.]
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#22 Writersblock

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Posted 24 November 2007 - 07:25 AM

Yeah, weird. 26 should have some significance. Here's a puzzler I ran across. Is this proveable?

N=p1+p2 where N>2.
Where N is a rational whole number, P1 is a prime, P2 is a prime not equal to P1.
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#23 bonanova

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Posted 24 November 2007 - 01:26 PM

Yeah, weird. 26 should have some significance. Here's a puzzler I ran across. Is this proveable?

N=p1+p2 where N>2.
Where N is a rational whole number, P1 is a prime, P2 is a prime not equal to P1.

Well no, because primes are odd. So N couldn't be odd.

Goldbach suggested N = p1 + p2 + p3 when N is odd.
Let p3=1, then [N-p3] = p1 + p2.
Since N-p3 is now even, they're essentially the same conjecture.

Neither version has been proved - although everyone believes they're true.
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#24 bonanova

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Posted 24 November 2007 - 02:09 PM

I just checked and learned that 1 is not considered a prime number [3 is the smallest odd prime]
so Goldbach's "weak" conjecture applies for odd N greater than 5.

N = p1 + p2 + p3; [N odd, >5]

Let p3=3
[N-3] = p1 + p2; [(N-3) is even, >2]
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#25 spoxjox

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Posted 06 December 2007 - 10:07 PM

I just checked and learned that 1 is not considered a prime number [3 is the smallest odd prime]
so Goldbach's "weak" conjecture applies for odd N greater than 5.

N = p1 + p2 + p3; [N odd, >5]

Let p3=3
[N-3] = p1 + p2; [(N-3) is even, >2]



Considering 1 not to be prime, this conjecture fails for 11.
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#26 spoxjox

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Posted 06 December 2007 - 10:10 PM

Since 'a' is a prime number, without loss of generality we can consider it to be an odd number (the exception is 2, and it is clear that 2*2+26 is not a prime).
Then 'a' can be written as "a = 2b+1", for some 'b' a natural number.
a*a = (2b+1)*(2b+1) = 4b^2 + 4b + 1.
(a*a)+26 = 4b^2 + 4b + 1 + 26 = (4b*(b+1))+27 ...... (*)

But to show that (a*a)+26 is not prime, it suffices to show that (4b*(b+1))+27 is a multiple of three. Of course 27 is a multiple of three, so we focus on the second term. But then either 'b' or 'b+1' must be a multiple of three, which is equivalent to say that 'b-1' (OR 'b+2') is not a multiple of three. This is because, for any natural number X, one and only one of 'X', 'X+1', 'X+2' is divisible by three.

a = 2b+1
==> b = (a-1)/2
==> b - 1 = ((a-1)/2) - 1 = (a-3)/2 .... (**)
In order 'b-1' to be a multiple of three, 'a' must be a multiple of three -- which cannot be true because 'a' is prime. Therefore, either 'b' or 'b+1' must be a multiple of three, which implies (4b*(b+1))+27 is a multiple of three. And hence (a*a)+26 is a not prime.

If anyone noticed a flaw with the argument, it would be nice to post your comments.



I can't immediately spot the problem with your proof, but it must be flawed, because your same reasoning can apply to a*a + 8, which fails at a=3.
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#27 bonanova

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Posted 07 December 2007 - 03:39 AM


I just checked and learned that 1 is not considered a prime number [3 is the smallest odd prime]
so Goldbach's "weak" conjecture applies for odd N greater than 5.

N = p1 + p2 + p3; [N odd, >5]

Let p3=3
[N-3] = p1 + p2; [(N-3) is even, >2]



Considering 1 not to be prime, this conjecture fails for 11.


11 = 3 + 3 + 5. Are you objecting to using 3 twice?
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#28 spoxjox

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Posted 07 December 2007 - 04:04 AM



I just checked and learned that 1 is not considered a prime number [3 is the smallest odd prime]
so Goldbach's "weak" conjecture applies for odd N greater than 5.

N = p1 + p2 + p3; [N odd, >5]

Let p3=3
[N-3] = p1 + p2; [(N-3) is even, >2]



Considering 1 not to be prime, this conjecture fails for 11.


11 = 3 + 3 + 5. Are you objecting to using 3 twice?


Yes, I thought that was the condition, that p1, p2, and p3 were distinct primes. My apologies if I'm conflating two conversations.
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#29 brhan

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Posted 07 December 2007 - 12:06 PM


Since 'a' is a prime number, without loss of generality we can consider it to be an odd number (the exception is 2, and it is clear that 2*2+26 is not a prime).
Then 'a' can be written as "a = 2b+1", for some 'b' a natural number.
a*a = (2b+1)*(2b+1) = 4b^2 + 4b + 1.
(a*a)+26 = 4b^2 + 4b + 1 + 26 = (4b*(b+1))+27 ...... (*)

But to show that (a*a)+26 is not prime, it suffices to show that (4b*(b+1))+27 is a multiple of three. Of course 27 is a multiple of three, so we focus on the second term. But then either 'b' or 'b+1' must be a multiple of three, which is equivalent to say that 'b-1' (OR 'b+2') is not a multiple of three. This is because, for any natural number X, one and only one of 'X', 'X+1', 'X+2' is divisible by three.

a = 2b+1
==> b = (a-1)/2
==> b - 1 = ((a-1)/2) - 1 = (a-3)/2 .... (**)
In order 'b-1' to be a multiple of three, 'a' must be a multiple of three -- which cannot be true because 'a' is prime. Therefore, either 'b' or 'b+1' must be a multiple of three, which implies (4b*(b+1))+27 is a multiple of three. And hence (a*a)+26 is a not prime.

If anyone noticed a flaw with the argument, it would be nice to post your comments.



I can't immediately spot the problem with your proof, but it must be flawed, because your same reasoning can apply to a*a + 8, which fails at a=3.



I forgot to include the condition that 'a' should be greater than 3 so that it can be written in the form of '2b+1' for 'b' a natural number. This is because 'b-1' will be zero if you take a=3. And as we all know zero is a multiple of any number.

Even in the original puzzle, if you take a=3, a*a+26 will be 35 which is not a prime number but it is not a multiple of three either. In a sense you can say my argument fails. Thanks for pointing me the error.
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#30 bonanova

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Posted 08 December 2007 - 06:07 AM




Considering 1 not to be prime, this conjecture fails for 11.


11 = 3 + 3 + 5. Are you objecting to using 3 twice?


Yes, I thought that was the condition, that p1, p2, and p3 were distinct primes. My apologies if I'm conflating two conversations.


In the strong form, the primes are distinct; the weak form [three primes for odd numbers] relaxes that condition.
If the 3 can be repeated, then the weak form is established immediately from the strong form.
I'm not an expert in number theory. But I found this interesting - that it's universally believed but still not proved.

There's more about it here.
http://en.wikipedia.org/wiki/Goldbach's_weak_conjecture
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