prime failed proof. A re-examination clearly demonstrated the logical flaw.

However, I set about the puzzle with renewed vigour, and have discovered

that if (p,p+2) are twin primes then p+2 divides p.2^p+1. Here goes:

p.2^p+1 = (p+2).2^p - [ 2^(p+1) - 1]

If p+2 is a prime, then p+1 = p+2-1, hence 2^(p+1)-1 = 2^((p+2)-1)-1 ==

0mod(p+2)

Q.E.D.

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths/

http://www.users.globalnet.co.uk/~perry/DIVMenu/

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