Any prime number greater than or equal to 5 can be expressed in the form 6p+1 or 6p-1, where p is a positive integer. Can you prove it?

The proof is very simple.....

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Guest Message by DevFuse

Started by amlan, Jun 24 2009 06:18 AM

3 replies to this topic

### #1

Posted 24 June 2009 - 06:18 AM

### #2

Posted 24 June 2009 - 06:56 AM

The problem should read like this.... either 6p+1 or 6p-1..

e.g. 41 is in the form 6p-1 (where p is 7) and 13 is in the form 6p+1 (where p is 2)....

e.g. 41 is in the form 6p-1 (where p is 7) and 13 is in the form 6p+1 (where p is 2)....

### #3

Posted 24 June 2009 - 07:12 AM

Its easy one......

Any number can be written as 6P or 6P+1 or 6P+2 or 6P+3 or 6P+4 or 6P+5.........

Now, 6P, 6P+2 and 6P+4 are always divisible by 2. 6P+3 is always divisible by 3. so we are left with 6P+1 and 6P+5 for prime numbers......

And, 6P+5 is same as 6P-1.....

proved.

Any number can be written as 6P or 6P+1 or 6P+2 or 6P+3 or 6P+4 or 6P+5.........

Now, 6P, 6P+2 and 6P+4 are always divisible by 2. 6P+3 is always divisible by 3. so we are left with 6P+1 and 6P+5 for prime numbers......

And, 6P+5 is same as 6P-1.....

proved.

### #4

Posted 24 June 2009 - 07:14 AM

Any prime number greater than or equal to 5 can be expressed in the form 6p+1 or 6p-1, where p is a positive integer. Can you prove it?

The proof is very simple.....

any number can be expressed as a factor of 6 with r remainder:

x = 6m + r

where m is the multiple and

r >= 0 and r <=5 .

if r = 0 , 2 ,3, 4 the number cannot be prime since it will be divisible by 2 or 3

therfore for a number to be prime r must be 1 or 5.

Therefore

**x = 6m + 1**

or:

x = 6m +5 = 6(m+1) - 1 =

**6(new m) -1**

Thus proved

(Note all prime numbers will be contained in this... but not all such numbers will be prime.... this is a superset of prime numbers )

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