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Math - Prime number


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#31 Writersblock

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Posted 10 December 2007 - 06:26 AM

Yeah, all this kind of smokes me intellectually. I knew there was a reason I went to law school....
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#32 mangulo13

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Posted 10 December 2007 - 05:18 PM

I believe it is 7. 7*7= 49. 49+26= 75. correct? 75 isn't prime right? There might be more solutions but i stoped when I got one solution.
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#33 dhamma

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Posted 11 December 2007 - 07:42 AM



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.



It would appear from this that 4b(b+1)+1+x would give you a working case back to (a^2)+x if the following conditions are met:
x > 0 (just to be safe, this may not be necessary),
(x+1)%3 = 0 (e.g. 27 from the original)
2*2+x is not prime (since this case was proved by example)
3*3+x is not prime (this is needed because if (b-1)%3=0, then a%3=0; thus for a=2b+1 when (b-1)%3=0, a is not prime unless it is 3. This allows us to state that 4b(b+1) is divisible by 3 and therefore not prime as long as b > 1)

This appears sound to me right now, but it's starting to get late. Let me know if it checks out.
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#34 brhan

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Posted 11 December 2007 - 01:25 PM

It would appear from this that 4b(b+1)+1+x would give you a working case back to (a^2)+x if the following conditions are met:
x > 0 (just to be safe, this may not be necessary),
(x+1)%3 = 0 (e.g. 27 from the original)
2*2+x is not prime (since this case was proved by example)
3*3+x is not prime (this is needed because if (b-1)%3=0, then a%3=0; thus for a=2b+1 when (b-1)%3=0, a is not prime unless it is 3. This allows us to state that 4b(b+1) is divisible by 3 and therefore not prime as long as b > 1)

This appears sound to me right now, but it's starting to get late. Let me know if it checks out.



It is correct.
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#35 stanger828

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Posted 27 January 2008 - 10:42 PM

Thinking along the same lines as Cawebster, 3 IS the magic number. Here is a proof of the problem:


for all primes "a" except 3, a^2+26 is divisible by 3


Part I - Testing when a=2 and when a=3

let a=2
2^2+26= 30 = Composite

let a=3
3^2+26= 35 = Composite

Part II - All primes a except 3, a^2+26 is divisible by 3

When a number is divided by 3, the remainder can be either 0,1, or 2. As such, any positive integer greater than 2 can be written in the following forms:

m=some non-negative integer

3m
3m+1
3m+2

If a is a prime number unequal to three (we already solved the case when a=3 and a=2) then a can be written as 3m+1 or 3m+2.

The case when a=3m+1
a^2+26 = (3m+1)^2+26 = 9m^2+6m+27 = 3(3m^2+2m+9) Because this is divisible by three, then when a=3m+1 the result is COMPOSITE

The case when a=3m+2
a^2+26 = (3m+2)^2+26 = 9m^2+6m+30 = 3(3m^2+2m+10) Because this is divisible by three, then when a=3m+2 the result is COMPOSITE

QED
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#36 kevin1203

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Posted 01 February 2008 - 05:16 AM

The answer is: 2=a

2*2+26=30

30 factors are: 1,2,3,5,6,10,15,30

A prime number is only divisible by 1 and itself. 2 is the smallest and only even prime number. The only answer can be 2 otherwise it equals a prime number.
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#37 stanger828

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Posted 02 February 2008 - 01:00 PM

The answer is: 2=a

2*2+26=30

30 factors are: 1,2,3,5,6,10,15,30

A prime number is only divisible by 1 and itself. 2 is the smallest and only even prime number. The only answer can be 2 otherwise it equals a prime number.


If only it were that simple... but as I have demonstrated in my proof: 3*3+26= 35 and 35 is composite therefore the conjecture that the only answer is 2 is incorrect.
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#38 amateur

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Posted 05 March 2008 - 10:20 PM

**********************************
The theory is incorrect. a*a+26 can be prime if 'a' is prime. Ex when a=11, a*a+26 is 137 which is a prime number.

Edited by amateur, 05 March 2008 - 10:28 PM.

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#39 amateur

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Posted 05 March 2008 - 10:33 PM

**********************************
The theory is incorrect. a*a+26 can be prime if 'a' is prime. Ex when a=11, a*a+26 is 137 which is a prime number.


sorry my mistake 11*11+26 is 147 :)
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#40 brhan

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Posted 05 March 2008 - 11:10 PM

sorry my mistake 11*11+26 is 147 :)

.... and 147 is not prime. Therefore, the theory is correct, right? :)
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