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Math - Prime number


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#1 Vobnalb9

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Posted 17 November 2007 - 03:24 PM

If "a" is a prime number, then prove, that a*a+26 is NOT a prime number (whether it is true).
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#2 bonanova

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Posted 17 November 2007 - 03:26 PM

Clarification: [a*a]+26? -or- a*[a+26]?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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#3 Vobnalb9

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Posted 17 November 2007 - 03:48 PM

[a*a]+26
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#4 Writersblock

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Posted 18 November 2007 - 08:45 AM

Yes, to what bona wrote below. I removed this post as soon as I put it up as I realized it was wrong. I errantly said that you could factor 13 and 2 from every solution of (A*A)+26. This is obviously wrong. Let's see if I can think of something correct now...
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#5 bonanova

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Posted 18 November 2007 - 08:48 AM

It might be harder than that. If 26 were a multiple, but it's only added.
I'm at a loss on this one.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#6 Writersblock

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Posted 18 November 2007 - 09:09 AM

I think I might have it, straining hard to remember high school algebra and primes. Ok... all primes above ... seven I think ... can be notated as 6n+1 or 6n-1. If you square these, then you have 36n+12n+1 and 36n-12n+1. Right? So you factor these down and you get 12n(3n+1) and 12n(3n-1).

Umm... brain is smoking now, but let's keep at it....

So, take 12n(3n+1). IF n is odd, then 3n+1 must be even, if 3n+1 is odd, then n must be even. So, 12 times an even number is divisible by 24, so 24 is a factor of every prime over 7 (notated as 6n+1) when it is squared. If you add 24 to that, you still have a non-prime. OK?

So, take 12n(3n-1). Ah HA! Same thing. N or 3n-1 must be even, so 24 will be a factor of every prime notated as 6n-1 so we have the same result.

So now we just have to figure out primes less than 7.

7: 49+24 = 73 = PRIME! DAMN! Should have done that first ...

5: 25+24 = 69 = not prime.

3: 9+24 = 33 = not prime.

1: 25 = not prime.

Seven proves it's not true. But, for all primes except for 7 it's true.
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#7 Writersblock

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Posted 18 November 2007 - 09:13 AM

*Sigh*

Now I know why I never became a mathmatician....

It was +26 not +24.

Let's see what else we can come up with.
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#8 Writersblock

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Posted 18 November 2007 - 09:19 AM

ARG! I messed up the multiplication of the binomials too I think. I'm too tired right now, so I'll start this over later on when I am fresh!
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#9 Writersblock

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Posted 18 November 2007 - 07:58 PM

Hmmm... Is it this easy? The square of any two primes over 7 (if I remember correctly) can be notated as either (6n+1)^2 or (6n-1)^2
This gives 36n^2+12n+1 and 36n^2-12n+1. Both have the +1 as part of it, so if you add 26 then both will have +27.

So using the first you have 36n^2+12n+27 and in the second you have 36n^2-12n+27. Both of these factor by 3, so therefore no prime over 7 when squared and added to 26 can equal a prime.

I just realized that the 6n+1 notation is good for 7 also. As is the 6n-1 for 5.

So 3,2,&1:

(3^2)+26 = 35 Non prime
(2^2)+26 = 30 Non prime
27 = Non Prime

Therefore where A = Prime, A^2+26=Prime is false.
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#10 cawebster

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Posted 19 November 2007 - 06:49 AM

Hmmm... Is it this easy? The square of any two primes over 7 (if I remember correctly) can be notated as either (6n+1)^2 or (6n-1)^2
This gives 36n^2+12n+1 and 36n^2-12n+1. Both have the +1 as part of it, so if you add 26 then both will have +27.

So using the first you have 36n^2+12n+27 and in the second you have 36n^2-12n+27. Both of these factor by 3, so therefore no prime over 7 when squared and added to 26 can equal a prime.

I just realized that the 6n+1 notation is good for 7 also. As is the 6n-1 for 5.

So 3,2,&1:

(3^2)+26 = 35 Non prime
(2^2)+26 = 30 Non prime
27 = Non Prime

Therefore where A = Prime, A^2+26=Prime is false.



This is a long explanation. I tried to be thorough. Let me know if you find any flaws.

Consider the set of perfect squares:
{ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, ... }

0+1 = 1
1+3 = 4
4+5 = 9
9+7 = 16

In short let B0 represent the first element in the sequence 0,
let B1 represent the second element, and Bn represent the (n-1)th element in the sequence.

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

Using this rule for the sequence of perfect squares we can now see that a pattern develops with regard to the remainders after division by 6. (I must credit WritersBlock here. I've never heard of the rule about sqares of primes expressed as 6n+1. I never would have looked at the remainders as a point of interest had I not seen the previous post.)

0/6 = 0 r 0-----1/6 = 0 r 1-----4/6 = 0 r 4-----9/6 = 1 r 3-----16/6 = 2 r 4------25/6 = 4 r 1
36/6 = 6 r 0---49/6 = 8 r 1----64/6 = 10 r 4---81/6 = 13 r 3---100/6 = 16 r 4--121/6 = 20 r 1
...

This pattern is guaranteed to continue because we're adding consecutive odd numbers and every six consecutive odd numbers will add a multiple of six. So, whatever the remainder is for any given number, it will be the same for the sixth number after it.
Proof: Let k represent any odd number.
add k and the 5 consecutive odd numbers after k.
k + k+2 + k+4 + k+6 + k+8 + k+10 =
6k + 30 = 6(k+5)

Using the pattern of remainders {0, 1, 4, 3, 4, 1}, only the second and fifth columns could possibly be the sqares of prime numbers.
If remainder is 0 then A^2 is divisible by 6 which means that A is not prime.
-----(if A is prime then the prime factors of A^2 is A*A. 6 can't be a factor)
If remainder is 4 then A^2 is divisible by 2 which means that A is 2 or "not Prime".
-----(if A is prime then the prime factors of A^2 is A*A. 2 can't be a factor unless A=2)
If remainder is 3 then A^2 is divisible by 3 which means that A is 3 or "not Prime".
-----(if A is prime then the prime factors of A^2 is A*A. 3 can't be a factor unless A=3)
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.
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