## Friday, December 26, 2008

### Twin Primes

First off, a twin prime is a set of two consecutive odd numbers both of which are primes.

Most mathematicians believe that the twin prime series is an infinite one. The conjecture though, has never been proven.
I came up with a simple proof by method of contradiction.

Consider the twin prime series to be S
S = (3,5) , (11,13) , (17,19) , (29,31) ..... (N-1, N+1)
Where (N-1, N+1) is the last twin prime.

Take the product of all the primes till (N+1) to get a number, P.
P = 1 x 2 x 3 x 5 x 7 x 11 x .... x (N-1) x (N+1)

(P+1) will not be divisible by any prime number, making it a prime.

Similarly, (P-1) won't be divisible by any prime. Making it a prime as well.

Therefore, (P+1, P-1) is a twin prime.
And by contradiction, (N+1, N-1) is not the final twin prime.

Conclusively, the series is infinite.

Now, as I said the argument is very simple and it'd be ludicrous to think I'm the first to think of it. Which further means there is a hole or error in it somewhere. So, if you spot anything, help me out.

Markkimarkkonnen said...

Nikita,

It's a tempting argument, but unfortunately there is an error. Your numbers "P+1" and "P-1" are not necessarily prime.

Although they are not divisible by any numbers <= (N+1), there could still be other primes, which are greater than (N+1), that could divide either (P-1) or (P+1).

For example, if I claim the only twin primes are (3,5) and (5,7), your method would attempt to find a contradiction with my claim by calculating
P = 2*3*5*7 = 210.

Then it would claim that 209 and 211 are prime, but in fact 209 = 11*19.

You might take a shot at proving the following theorems about twin primes:

1) The only number between a pair of twin primes that is not divisible by six is 4. (i.e. (11,13) is a twin prime, 12 is in between, and it is a multiple of 6. Likewise for 30, which is between the twin primes (29,30).)

2) The only "triplet primes", that is, three numbers, all prime, each separated by two are (3,5,7).

Nik said...

Ah. Thank you.

Thanks for the theorems too. Got em both. Pretty simple. I'm certain I have it right this time.