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# Math - Prime number

### #41

Posted 05 March 2008 - 11:51 PM

did I get it right?

### #42

Posted 06 March 2008 - 12:02 AM

### #43

Posted 06 March 2008 - 06:38 AM

Any number (a) times itself creates an even number added to an even number (26) will always be divisible by 2, by meaning NOT PRIME.

did I get it right?

Ny number times itself is even???? like..3*3=9 (even?), 5*5=25(even?),......your math is off i think

### #44

Posted 06 March 2008 - 11:19 PM

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.

First, we know that 3 does not divide 'a' because 'a' is prime. Then,

Suppose the prime 'a' is congruent to 1 mod 3. We can write 'a' as 3n + 1 where 'n' is a natural number (integer > 0).

Then (3n + 1)*(3n + 1)=9*(n^2) + 6n + 1 using the FOIL method. It is easy to see that '9*(n^2)' can be divided by 3 as can '6n'.

But 1 is still congruent to 1 mod 3.

This means we can rewrite the sum as a*a=3k + 1 where 'k' is a natural number.

So if 'a' is congruent to 1 mod 3, so is 'a*a'.

Then,

Suppose the prime 'a' is congruent to 2 mod 3. We can write 'a' as 3n + 2 where 'n' is a natural number (integer > 0).

Then (3n + 2)*(3n + 2)=9*(n^2) + 12n + 4 using the FOIL method. It is easy to see that '9*(n^2)' can be divided by 3 as can '12n'.

This is the same as before, but with one extra step. We see that 4 is congruent to 1 mod 3.

This means we can rewrite the sum as a*a=3k + 1 where 'k' is a natural number, even if 'a' is congruent to 2 mod 3!

In conclusion, 'a*a' always has a remainder of 1 when divided by 3, and 26 has a remainder of 2 when divided by 3.

This means when the remainders are added together, they are ALWAYS divisible by three, thus it cannot be a prime.

This method works for any number like 26, where it is congruent to 2 mod 3.

So 'a*a + 2', 'a*a + 5', 'a*a + 212' are all not primes, as long as 'a' is prime.

-Mathematics Graduate Student

University of Missouri

### #45

Posted 07 March 2008 - 12:50 PM

In conclusion, 'a*a' always has a remainder of 1 when divided by 3, and 26 has a remainder of 2 when divided by 3.

This means when the remainders are added together, they are ALWAYS divisible by three, thus it cannot be a prime.

This method works for any number like 26, where it is congruent to 2 mod 3.

So 'a*a + 2', 'a*a + 5', 'a*a + 212' are all not primes, as long as 'a' is prime.

Correct, with one condition though. 'a' should be greater than three. The reason can be found in the previous threads.

For a=3, 3*3+26=35 which is not divisible by three despite 'a' being a prime number.

### #46

Posted 16 March 2008 - 03:52 AM

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

Me being who I am, I took the liberty of writing out some of the equations

Prime numbers -- Soultion-Prime number factor

2 -- 30 Composite -(2)

3 -- 35 Composite - (5)

5 -- 51 Composite - (3)

7 -- 74 Composite - (2)

11 -- 147 Composite - (3)

13 -- 195 Composite - (3)

17 -- 315 Composite- (5)

19 -- 387 Composite- (5)

Unless otherwise, I believe that all prime numbers squared, plus 26 is a composite number.

Thus concluding that the answer would be:

**Edited by Standards, 16 March 2008 - 03:56 AM.**

### #47

Posted 16 March 2008 - 04:58 PM

Me being who I am, I took the liberty of writing out some of the equations

Prime numbers -- Soultion-Prime number factor

2 -- 30 Composite -(2)

3 -- 35 Composite - (5)

5 -- 51 Composite - (3)

7 -- 74 Composite - (2)

11 -- 147 Composite - (3)

13 -- 195 Composite - (3)

17 -- 315 Composite- (5)

19 -- 387 Composite- (5)

Unless otherwise, I believe that all prime numbers squared, plus 26 is a composite number.

Thus concluding that the answer would be All prime numbers.

It is true that if "a" is a prime number, then "a*a+26" is not a prime. But the method of proof you followed is not correct. Look at the following conjecture.

For any positive integer 6n-1 or 6n+1 (or both) is a prime.

According to your proof method:

1 -- 5 Prime (6n-1)

2 -- 11 Prime (6n-1)

3 -- 17 Prime (6n-1)

4 -- 23 Prime (6n-1)

5 -- 29 Prime (6n-1)

6 -- 37 Prime (6n+1) ... here 6n-1 is 35 which is a composite number.

7 -- 41 Prime (6n-1)

8 -- 47 Prime (6n-1)

But the conjecture is not correct. Take n=20.

6*20+1 = 121, which is divisible by 11.

6*20-1 = 119, which is divisible by 7 (and 17).

So, I suggest you better use one of the well known mathematical proof methods.

### #48

Posted 16 March 2008 - 06:50 PM

### #49

Posted 16 March 2008 - 07:03 PM

I know, and said that in my post. But I am commenting on YOUR PROOF METHOD. I cannot proof a conjecture by exhaustion because there are infinite numbers, and it is impossible to test all of them. I was just trying to demonstrate such kind of proof method doesn't want all the time.But, we're trying to prove how the expression (a*a)+26 can not equal another prime number, not how we find a prime number.

### #50

Posted 19 March 2008 - 04:10 AM

I know, and said that in my post. But I am commenting on YOUR PROOF METHOD. I cannot proof a conjecture by exhaustion because there are infinite numbers, and it is impossible to test all of them. I was just trying to demonstrate such kind of proof method doesn't want all the time.

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

But other than that I understand.

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