## Extended conjecture

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• twin prim conjecture extended Posted by: Sebastian Martin sebi_sebi@yahoo.com sebi_sebi Date: Thu Jul 6, 2006 4:32 pm (PDT) Hello I have extended the
Message 1 of 1 , Jul 8, 2006
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twin prim conjecture extended
Posted by: "Sebastian Martin" sebi_sebi@... sebi_sebi
Date: Thu Jul 6, 2006 4:32 pm (PDT)

Hello

I have extended the conjecture:

p, q consecutive primes p<q

are equivalents:

1) p q are twin primes

2) z=sqrt[(p^2+q^2)/2-1] is integer

3) w=sqrt[p*q+1] is integer

considering 2) ---> 3) where q and p are consecutive primes.

Suppose

z^2 = [p^2 + q^2 ] /2 - 1

(q-p)^2 = p^2 + q^2 - 2 p q

(q-p)^2 /2 = [ p^2 + q^2 ] /2 - p q

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

(q-p)^2 /2 - 1 = z^2 - pq

(q-p)^2 / 2 - 1 - 1 = z^2 - pq - 1

(q - p)^2 / 2- 2 = z^2 - [ pq + 1 ]

(q-p)^2 / 2 = z^2 - [pq + 1 ] + 2

(q - p)^2 = 2 ( z^2 - [ pq + 1 ] ) + 4

That is probably what you did to get your extension of the conjecture.

We still need to see why z^2 = pq + 1,

where z^2 = [ p^2 + q^2 ]/2 - 1,

and

we know only that q is the next prime after p.

If we know that z^2 = pq + 1, then we can prove, q = p + 2.

(q - p)^2 = 2 ( z^2 - [ pq + 1 ] ) + 4

That is, you have shown that

if q is the next prime after p, and

z^2 = [ p^2 + q^2]/2 - 1 = pq + 1

then q = p + 2.

This is not quite your original conjecture, but is now
cast in a provable form.

Kermit < kermit@... >

Kermit < kermit@... >
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