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## twin primes conjecture: A property of twin primes, only?

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• Re: Discussion between Sebastian Martin Ruiz and Mike Oakes ... sqrt([ (23^2 + 25^2)/2 -1) = sqrt(529 + 625)/2 - 1) = sqrt(1154/2 -1) =sqrt(577 - 1) =
Message 1 of 1 , Jul 4, 2006
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Re: Discussion between

Sebastian Martin Ruiz
and
Mike Oakes

> Let p q consecutive prime numbers p<q.
> Let z=sqrt[(p^2+q^2)/2-1]
>

> Conjecture: p&q ares twin primes IF AND ONLY IF z is
> integer.

sqrt([ (23^2 + 25^2)/2 -1)

= sqrt(529 + 625)/2 - 1)
= sqrt(1154/2 -1)
=sqrt(577 - 1)
= sqrt(576)
= 24

Is this a counter example to the if part?

The only if part can be easily proven,
for it follows from the fact that of two twin primes,
p,q with p < q, that

p is of the form 6 D -1 and q is of the form 6 D + 1.

[ Except for the twin primes 3 and 5 ]

Let D be a positive integer that is not any of the four
forms

6 m n - m - n
6 m n - m + n
6 m n + m - n
6 m n + m + n

with m > 0, n > 0.

Then both 6D -1 and 6 D +1 are twin primes,

and all sets of twin primes are of this form.

D = 1 is permitted because 1 is not any of the four forms.

4 = 6 - 1 - 1 is the smallest positive integer of any of these forms.

So D= 1 yields the twin primes 5 and 7.
D = 2 yields the twin primes 11 and 13.
D = 3 yields the twin primes 17 and 19.

Because 4 = 6 - 1 -1,

23 and 25 are not twin primes.

Let p and q be twin primes,
p > 3.

Set p = 6 D -1
q = 6 D + 1

[ p^2 + q^2 ] = [ 36 D^2 - 12 D + 1 ] + [ 36 D^2 + 12 D + 1 ]
= 72 D^2 + 2

[ p^2 + q^2 ] /2 = 36 D^2 + 1

sqrt[(p^2+q^2)/2-1] = 6 D = p + 1
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