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# Math - Prime number

## 88 posts in this topic

Posted · Report post

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

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Posted · Report post

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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Posted · Report post

"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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Posted · Report post

"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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Posted · Report post

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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Posted · Report post

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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Posted · Report post

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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Posted (edited) · Report post

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
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Posted (edited) · Report post

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
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Posted · Report post

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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Posted (edited) · Report post

I just saw this puzzle today and see it a bit differently.

We all know that 2 is the only even prime number.

We also know that an odd number times an odd number is always an even number. The same is true for multiplying even numbers (even x even = even)

So...

(A*A)+26 can be seen as (ODD*ODD)+EVEN or (EVEN*EVEN)+EVEN. Either way the result is always even and always greater than 2. As such, this expression can never yield a prime result. The answer will always be non-prime.

Edited by Wizzbang
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Posted (edited) · Report post

We also know that an odd number times an odd number is always an even number.

I have to disagree. An odd number, say 5 times 5.. is not an even number...

5*5=25

also

7*7=49

last time i checked, 25 and 49 were not even numbers...

im pretty sure its all odd numbers times an odd number equals an odd number and all even numbers times an even number equals an even number.

Edited by cmabb21
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Posted (edited) · Report post

if a was any number, and you squared it, it is guaranteed to have divisible factors, making it composite(i.e 11*11=121 with factors 1, 11, 121. 7*7=49 with factors 1, 7, 49. 5*5=25 with factors 1, 5, 25). So even if a*a is an odd number there will be at least 3 factors to it making it a composite number and not prime. if a happened to be 1, then 1*1+26 would be 27 which is a composite number with factors 1,3,9, and 27.

Edited by toumba
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Posted · Report post

Sorry, I stumbled upon this one so late.

Here is a SIMPLE PROOF:

Any prime number P may be expressed as P=3*n + 1 or P=3*n+2. (Another way of saying prime gives a remainder of 2 or 1, when divided by 3).

Thus P2 + 26 can be expressed as 9n2+6n+27 or 9n2+12n+30. Both expressions are wholly divisible by 3.

Therefore, P2 + 26 is always divisible by 3.

QED

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Posted · Report post

Any prime number P may be expressed as P=3*n + 1 or P=3*n+2. (Another way of saying prime gives a remainder of 2 or 1, when divided by 3).

Technical correction: Any prime number, but 3 -- that is. For 3 we have to check separately: 32 + 26 = 35 -- not a prime. Everyting else is divisible by 3.

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Posted · Report post

Hey all, I just skimmed through this topic and saw some pretty complicated math. While quite a few people got to the correct answer I have a way that I think is a good bit simpler than anything I saw. So, just in case anybody is interestedâ€¦

(A*A) + 26 is never prime if A is prime.

The key to my way is modular arithmetic. Sense, A is prime we know that it is not divisible by 3. Therefore, A modulo 3 = 1 or 2. (This means that A / 3 has a remainder of 1 or 2.) Letâ€™s split this into two cases.

*It is important to know that modular arithmetic works the same as regular arithmetic when dealing with addition and multiplication.

Case 1: A modulo 3 = 1: Thus A*A modulo 3 = 1*1 = 1

Case 2: A modulo 3 = 2: Thus A*A modulo 3 = 2*2 = 4 = 1 modulo 3**

**Here I used the fact that 4 and 1 are equal with respect to modulo 3. That is because both of them are 1 more than a multiple of 3. This is the definition of modulo arithmetic.

From the two above cases we have proved that any A*A = 1 modulo 3 when A is prime. Now by adding any number that equals 2 modulo 3 (which is equal to -1 modulo 3; they are both 1 less than a multiple of 3) we see that we will get a number that equals 0 modulo 3. In other words the number will be divisible by 3 and thus not prime! I saw bonanova touched on this, 26 is completely arbitrary. Any number that equals -1 modulo 3 will work.

Sorry, I didnâ€™t explain this very well, but it really is quite simple when you think of it in these terms. (And really sorry if someone already gave this process I just skimmed the first few pages. Hmmm... and this is basically what prime just said, but oh well I typed it all out now. He is right though 3 will have to be tested seperately, but since 35 is not a prime we're good.)

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Posted · Report post

a=3

3*3=9

9+26=35

35 is not prime

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Posted · Report post

Showing that a*a+26 is not prime when a=3 is true but not a math. proof. However it is quite easy: (if I didn't misunderstand)

a*a + 26 =

a*a -1 + 27=

(a-1)*(a+1) + 27

Now, since a is prime, it can not be a mutliplier of 3. Then either (a+1) or (a-1) should be a multiplier of 3.

Then (a-1)*(a+1) is multiplier of 3

And (a-1)*(a+1) +27 is also multiplier of 3

Then it's not prime.

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Posted · Report post

It might be harder than that. If 26 were a multiple, but it's only added.

I'm at a loss on this one.

I'm not sure if your equation is right but theres an easier way. the smallest non prime number that is undebateable (1 is debateable) is 4.

4*4 is the same as 4x4x4x4 which is 256. add 26 to that and you have 282 which can be divided by 2 and thus not prime.

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Posted · Report post

I'm not sure if your equation is right but theres an easier way. the smallest non prime number that is undebateable (1 is debateable) is 4.

4*4 is the same as 4x4x4x4 which is 256. add 26 to that and you have 282 which can be divided by 2 and thus not prime.

Well, 4^4 = 4 x 4 x 4 x 4

But 4*4 = 4 x 4

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Posted (edited) · Report post

Assume [a*a]+26=X

[a^2]+26=X

If a is: -> X is:

2 -> 30

3 -> 35

5 -> 51

7 -> 75

... -> ...

So why does <a> come out composite?

Quite simple really:

a^2 always has more factors, no matter what a is (given that it is prime).

Adding 26 is just a red herring in the problem.

Edited by Royal
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Posted · Report post

Assume [a*a]+26=X

[a^2]+26=X

If a is: -> X is:

2 -> 30

3 -> 35

5 -> 51

7 -> 75

... -> ...

So why does <a> come out composite?

Quite simple really:

a^2 always has more factors, no matter what a is (given that it is prime).

Adding 26 is just a red herring in the problem.

Why 26 is a herring?

Then proof the same problem when 26 is 27.

I mean proof that a*a + 27 is not prime when a is prime.

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Posted · Report post

...

Now, since a is prime, it can not be a mutliplier of 3. Then either (a+1) or (a-1) should be a multiplier of 3.

Then (a-1)*(a+1) is multiplier of 3

And (a-1)*(a+1) +27 is also multiplier of 3

Then it's not prime.

...

3 is a prime, yet it is a multiplier of 3.

(3-1)*(3+1) is not a multiplier of 3.

(3-1)*(3+1) + 27 is also not a multiplier of 3.

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Posted · Report post

Assume [a*a]+26=X

[a^2]+26=X

If a is: -> X is:

2 -> 30

3 -> 35

5 -> 51

7 -> 75

... -> ...

So why does <a> come out composite?

Quite simple really:

a^2 always has more factors, no matter what a is (given that it is prime).

Adding 26 is just a red herring in the problem.

Try P2 + 40. Where P is a prime and P > 5.

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Posted (edited) · Report post

3 is a prime, yet it is a multiplier of 3.

(3-1)*(3+1) is not a multiplier of 3.

(3-1)*(3+1) + 27 is also not a multiplier of 3.

You're right, but this doesn't make my proof invalid.

I should only make a modification:

Now, since a is prime, IF IT IS NOT 3 THEN it can not be a mutliplier of 3. Then either (a+1) or (a-1) should be a multiplier of 3.

IF IT IS 3 THEN 3*3 + 26= 35 and it is not prime. Then in all conditions, if a is prime a*a+26 is not prime.

I'll work on p^2 + 40

I've worked: 7^2 + 40 = 89

then it can be both prime and not.

Edited by nobody
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