## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Math - Prime number

87 replies to this topic

### #51 dinky

dinky

Newbie

• Members
• 2 posts

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

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
• 0

### #52 brhan

brhan

• Members
• 466 posts

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.
• 0

### #53 Standards

Standards

Newbie

• Members
• 19 posts

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.
• 0

### #54 brhan

brhan

• Members
• 466 posts

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.
• 0

### #55 dinky

dinky

Newbie

• Members
• 2 posts

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.
• 0

### #56 Fred

Fred

Newbie

• Members
• 1 posts

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
• 0

### #57 brhan

brhan

• Members
• 466 posts

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.
• 0

### #58 justme

justme

Newbie

• Members
• 1 posts

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.

• 0

### #59 Chuck

Chuck

Junior Member

• Members
• 78 posts

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.

• 0

### #60 cmabb21

cmabb21

Junior Member

• Members
• 23 posts

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.
• 0

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users