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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 127

p=any number ending in 3

check p is not a square

where a+b is a factor and a-b is another factor

if p is prime you will not find any factors

if p is composite you will find factors.

Example:

p=33

27*33=891

46+35=81=27*3 46-35=11

factors for 33 are 3 and 11

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,018

This can be disproved. if a and b are integers a+b and a-b are either both odd or both even. Therefore the product a^2-b^2 is either odd or divisible by 4. so if p is only divisible by 2 say 10 it is not possible to find a and b.( I mean 27p can be factorized into 2 odd or 2 even numbers.)

*Last edited by thickhead (2016-04-16 19:56:35)*

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 127

Sorry, this theorem only works with **ODD **numbers. So p is odd and a is either odd or even and b is the opposite.

Another way of proving p is prime is;

only when a-b=1So add possible to p and see if it = a square.

(a+b) and (a-b) will be factors

This is because for any odd Non-Prime that has a factor of >1, it can be produced with the below formula;

I can prove it

x = (f + L)/2

y = (f - L)/2 therefore

minus the two and it = fL which are the two factors....

And x - y = ((f + L)/2) - ((f - L)/2) = L which must be >1 for p to Not be, Prime or 1.

*Last edited by Primenumbers (2016-04-22 21:08:56)*

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,018

I am sorry. I had not noticed that p is a number ending in 3.

PRIME NUMBERS WROTE

" p= any prime number ending with 3. Check p is not a square." This is redundant. No square will end with 3.

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 127

Of course p does have those possible squares. But ONLY those. To see if p is composite, test all values for b where b is greater than 1and smaller than (p - 1)/2.

**"Time not important. Only life important."*** - The Fifth Element 1997*

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 127

Nehushtan wrote:

I don't really know how to do this. 4k+2 will always be even so n will be any odd no. My formula doesn't work for even no.'s. So n has factors f and L where a=(f+L) /2 and b=(f-L) /2. For a prime the only factors will be 1 and itself, so the only occurrence for a squared minus b squared will be when a=(p+1)/2 and b=(p-1)/2. For a composite b will be less than this.

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 127

Primenumbers wrote:

For a prime the only factors will be 1 and itself, so the only occurrence for a squared minus b squared will be when a=(p+1)/2 and b=(p-1)/2. For a composite b will be less than this.

**"Time not important. Only life important."*** - The Fifth Element 1997*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,663

Hi;

I have received two complaints about this thread and that is three too many.

Sir Isaac Newton wrote:

Tact is the art of making a point without making an enemy.

@Nehushtan: Please try to think of the other guys feelings when you respond. No more "rubbish" comments.

@Primenumbers: Please try to be more receptive when someone points out a possible error. Supposedly you are posting so that you can get feedback on your work. This is valuable to us all. Welcome criticism when you get it, this is how we all learn.

So I say, calm down everyone. Everyone knows the rules and everyone knows me.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**googol****Member**- From: Delft, The Netherlands
- Registered: 2016-04-22
- Posts: 13

Every composite number is the difference of two squares, a prime is never:

In your example

**10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,663

@Nehushtan

Did you read post #11? Using Einstein to demean a member is clever but still not allowed. Please stop deleting your posts and my additions in those posts.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 127

Primenumbers wrote:

p=any number ending in 3

check p is not a square

where a+b is a factor and a-b is another factor

Number ending in 3: factors end in 1 and 3, or 7 and 9. Times by 27 factors now end in 1 and 1.

f=10c+1 L=10d+1

b=(10c+1-10d-1)/2=5(c-d)

*Last edited by Primenumbers (2016-04-25 04:37:54)*

**"Time not important. Only life important."*** - The Fifth Element 1997*

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