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Math - Prime number

87 replies to this topic

#11 cawebster

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Posted 19 November 2007 - 07:04 AM

I messed up the sequence info slightly. It should read:

Consider the set of perfect squares:
{ 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, ... }

1+3 = 4
4+5 = 9
9+7 = 16

In short let B1 represent the first element in the sequence 0,
let B2 represent the second element, and Bn represent the (n)th element in the sequence.

n^2 = Bn = B(n-1) + 2n -1
(This can be proven by induction)

_______________________________________________________________________________
This does not change the logic for the rest of the problem. This is still valid for the proof.
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#12 brhan

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Posted 19 November 2007 - 12:30 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.
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#13 bonanova

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Posted 19 November 2007 - 04:06 PM

I was unaware of writersblock's 6n+/-1 formulas, but it looks as if it's true for any odd a,
thus for any 6n=/-1 and for any prime.

I wonder why 26 was chosen in the initial problem.
It seems that 26 could be replaced by any number that's 1 less than a multiple of 3. For example, -1.
But that would be trivial: [a*a]-1 is obviously even for any odd a.
So maybe 26 is just a number that looks arbitrary and would seem difficult to include in a proof.

Nice job, both!
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#14 Writersblock

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Posted 19 November 2007 - 09:27 PM

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 ...... (*)

This isn't true. We establish that A must be prime. 2b+1 = prime is not a true statement. Use b=4 for example.

Even though it's true that primes must be odd, it's not true that odds must be prime. You are committing a pretty basic error in logic by reassigning the question to fit your answer.
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#15 Writersblock

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Posted 19 November 2007 - 09:35 PM

Just don't use the definition of odd (2n+1 or 2n-1) use the definition for prime (6n+1 or 6n-1). I think your proof works then.

If you don't see how 6n+/-1 works, take a look at any prime number (you can find lists of them on the internet) and apply one or other of the formulas. It's always true.

For example, a large random prime = 47599
6n+1 doesn't work, but 6n-1 does.
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#16 Writersblock

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Posted 19 November 2007 - 09:57 PM

[quote="cawebster":2b940]
...
Therefore, if prime numbers exist (and we know they do) then their squares can be expressed as some multiple of six plus 1. Or, A^2 = 6x + 1

A^2 = 6x + 1
A^2 + 26 = 6x + 1 + 26
= 6x + 27
= 3(2x + 9) Which is obviously divisible by 3 and therefore not prime.
[/quote]

As to
[quote]
A^2= 6x + 1
[/quote] [/quote]
This should be A^2 = (6x+1)^2. And this is only true for some primes. Others are 6x-1.
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#17 bonanova

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Posted 20 November 2007 - 09:24 AM

Just don't use the definition of odd (2n+1 or 2n-1) use the definition for prime (6n+1 or 6n-1). I think your proof works then.

The proof works anyway, doesn't it? [primes are odd.]

Proving (a*a) + 26 is not prime whenever a is odd
may be shooting a rabbit with an elephant gun,
but the rabbit bites the dust regardless.
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#18 brhan

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Posted 20 November 2007 - 11:41 AM

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 ...... (*)

This isn't true. We establish that A must be prime. 2b+1 = prime is not a true statement. Use b=4 for example.

Even though it's true that primes must be odd, it's not true that odds must be prime. You are committing a pretty basic error in logic by reassigning the question to fit your answer.

Writersblock, nobody said all odd numbers are primes. All I said was all primes are odd (except 2.) As a result, any prime number (with the exception of 2) can written as '2b+1' or '2b-1' for some natural number 'b'. This is always true. The only exception, which is prime and even, is 2 -- and as I pointed out 2*2+26 is not prime.
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#19 brhan

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Posted 23 November 2007 - 11:29 AM

Just don't use the definition of odd (2n+1 or 2n-1) use the definition for prime (6n+1 or 6n-1). I think your proof works then.

If you don't see how 6n+/-1 works, take a look at any prime number (you can find lists of them on the internet) and apply one or other of the formulas. It's always true.

For example, a large random prime = 47599
6n+1 doesn't work, but 6n-1 does.

6n+/-1 is not a correct definition of prime numbers. Take n=20.
6*20+1 = 121, which is divisible by 11.
6*20-1 = 119, which is divisible by 7 (and 17).

But I agree with your argument that, any prime number can be written as 6n+/-1. My argument is somehow similar in that 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.
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#20 Writersblock

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Posted 23 November 2007 - 09:40 PM

Yeah, I get what you are saying.
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