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Math - Prime number


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#51 dinky

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Posted 19 March 2008 - 06:25 AM

To do this proof, we need to show that for any prime number 'a', ((a*a) + 26)/3 is an integer. We do this by the modulus of each number.
26 is congruent to 2 mod 3. Which means if 26 is divided by 3, the remainder is 2.
It then suffices to show that a*a is congruent to 1 mod 3. Which means if a*a is divided by 3, the remainder is 1.

...... blah blah blah

-Mathematics Graduate Student
University of Missouri


Here is another way to look at this proof:

The proof requires bases. I will be working in base three (number set 0, 1, 2)
Here is a base 3 multiplication table (leaving out 0) 1*1=1, 1*2=2, 2*2=11.
So, any number divisible by 3 will end in 0 (base 3).
A prime number is not divisible by 3, so it will end in either 1 or 2.
When I square that prime number, it will end in 1, always. (see multiplication table).
So, I just need to add 2, or 3n+2 to the square to make it divisible by 3.

I can use 3n + 2 in the place of "26" to make this puzzle work. So that the addend is not obvious, n is even.

Therefore, the problem works for a*a+2 as well as a*a+38
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#52 brhan

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Posted 19 March 2008 - 11:02 AM

It's prove, not proof, in your third sentence. And work, instead of want, last sentence.

But other than that I understand.


Well, I think can say either "your prove doesn't work" or "your proof method doesn't work". Your second comment is accepted (it was just a typo). Thanks for the correction.
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#53 Standards

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Posted 19 March 2008 - 02:13 PM

"I cannot proof a conjecture by exhaustion because there are infinite numbers, and it is impossible to test all of them."

That was the sentence I was referring to. It's, "I cannot prove a conjecture...."

Just letting you know.
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#54 brhan

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Posted 19 March 2008 - 02:45 PM

"I cannot proof a conjecture by exhaustion because there are infinite numbers, and it is impossible to test all of them."

That was the sentence I was referring to. It's, "I cannot prove a conjecture...."

Just letting you know.


Ok, I got it now. Thanks once more for the correction, and I appreciate for your sincere comments.
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#55 dinky

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Posted 21 March 2008 - 12:16 AM

Here is another way to look at this proof:

The proof requires bases. I will be working in base three (number set 0, 1, 2)
Here is a base 3 multiplication table (leaving out 0) 1*1=1, 1*2=2, 2*2=11.
So, any number divisible by 3 will end in 0 (base 3).
A prime number is not divisible by 3, so it will end in either 1 or 2.
When I square that prime number, it will end in 1, always. (see multiplication table).
So, I just need to add 2, or 3n+2 to the square to make it divisible by 3.

I can use 3n + 2 in the place of "26" to make this puzzle work. So that the addend is not obvious, n is even.

Therefore, the problem works for a*a+2 as well as a*a+38


However, this does not work for primes which are a multiple of 3. There is one. It is 3.
For that, we rely on 3*3 = 9. If the last number in 3n+2 ends in 6, then 9+6 ends in 5 which is divisible by 5.

Thus, the 2-digit numbers which work for a*a + K are

K=26, K=56, and K=86.
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#56 Fred

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Posted 07 April 2008 - 04:02 PM

This is easier than it looks:
a*a + 26 is always an even number, so it is not a prime.

All prime number except 2 are odd numbers. So a*a will always be an even number. If you add 26 which is an even number, then a*a + 26 is even, then it is not prime.
If a = 2 then 2*2+26=30 which is not a prime
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#57 brhan

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Posted 07 April 2008 - 04:11 PM

This is easier than it looks:
a*a + 26 is always an even number, so it is not a prime.

All prime number except 2 are odd numbers. So a*a will always be an even number. If you add 26 which is an even number, then a*a + 26 is even, then it is not prime.
If a = 2 then 2*2+26=30 which is not a prime

Nope!! that is not correct. If 'a' is odd 'a*a' is always odd ... 3*3=9; 5*5=25; 7*7=49 etc.. If you add an even number to an odd number the result is odd -- unlike your statement.

So, your proof method doesn't work.
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#58 justme

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Posted 11 April 2008 - 12:52 AM

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



With the exception of a=3, if a=prime number, then (a*a)+26 will always be divisible by 3. If a=3, the equation would still result in a non-prime number, it just would not be divisible by 3. Therefore the square of 'a' (which will not be prime) plus twenty six (which is not prime) results in a sum not prime.
Just a guess

Edited by justme, 11 April 2008 - 12:55 AM.

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#59 Chuck

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Posted 07 June 2008 - 03:40 AM

Good granny...just use mod 3. It's a much simpler/shorter proof.

Since a is a prime, either

a) a = 1 (mod 3)
b) a = -1 (mod 3)

Then, a*a = a2 = 1 (mod 3).

a2 + 26 = 0 (mod 3), and is therefore divisible by 3 and not a prime.

QED

Edited by Chuck, 07 June 2008 - 03:42 AM.

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#60 cmabb21

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Posted 07 July 2008 - 07:14 PM

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

a few prime numbers- (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139)

2*2+26=30 - not prime
2*3+26=32 - not prime
2*5+26=36 - not prime

you can keep going though.. my guess is only a few will be prime (if any). if its not true all the time, then its not true.
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