Wednesday, March 23, 2005

Twin primes

For a moment I thought I had proved the twin primes conjecture...

The conjecture is that there are infinitely many prime pairs differing by 2. eg. 5,7 ; 11,13 etc.

I started with Euclid's proof of the infiniteness of primes.
Assume for a contradiction that there are finitely many primes. Without loss of generality, assume that there are exactly n primes p1,p2, ... pn in increasing order.
Then, (p1.p2...pn + 1) must be a prime since it has no prime factors, contradicting our assumption.

In the same way, (p1.p2...pn - 1) must also be a prime if n>1.

Doesn't that give us a pair of primes (p1.p2...pn - 1), (p1.p2...pn + 1) for every n that differ by 2?

No, it does not. Why?

3 comments:

Anonymous said...

3*5 + 1 = 16 and
3*5 - 1 = 14
and both are not primes.

but still stunning reasoning!

Anonymous said...

sorry i missed it.... it should be
2*3*5 + 1 = 31
2*3*5 - 1 = 29
and yes it's amazing. what's wrong then?

Unknown said...

Try 2*3*5*7 - 1 = 209 is divisible by 11.

The supposition that the last prime in the product (eg. 7 here) is the largest prime in the universe - is part of the reasoning. Therefore, the reasoning is incorrect in general.

In order to check if a number n is a prime, you need only check for factors between 2 and sqrt(n). In the number 2*3*5*7-1, we know that 2,3,5,7 cannot be factors, but nothing prevents primes 7 < p <= sqrt(209) from being factors.