Dear Sebastain,

Did you mean by p**2+4(q+r)* *2=4p(q+r) +1

this p^2 + 4(q+r)^2 = 4p(q+r) + 1 ???

if so, maybe you need to clarify that:

r or q can equal p

or

1 is prime

see below:

11^2 + 4(q+r)^2 - 44(q+r) + 1 = 0

let w = q+r

121 + 4w^2 - 44w - 1 = 0

120 + 4w^2 - 44w = 0

divide both sides by 4

30 + w^2 - 11w = 0

as a quadratic equation:

w^2 - 11w + 30 = 0

a = 1

b = -11

c = 30

w = [-b +- sqrt(b^2 - 4ac)] / 2a

w = [11 +- sqrt(-11^2 - 4*1*30)] / 2*1

w = [11 +- sqrt(121 - 120)] / 2

w = [11 +- sqrt(1)] / 2

w = [11 +- 1] / 2

w1 = 12 / 2 = 5

w2 = 10 / 2 = 10

so

either:

w1 = q1 + r1

12 = 1 + 11 or 5 + 7

but but q != p and r != p

and 1 is not 1 since 1899

so this option is out

or:

w2 = q2 + r2

10 = 3 + 7 or 5 + 5

but q != p and r != p

and r != q

so this fails too unless you calrify that they can :)

Sorry if misunderstood ** and * * oparators.

Ali

<prime numbers are God's signature>

________________________________

From: Sebastian Martin Ruiz <

s_m_ruiz@...>

To: Claudi Alsina <

claudio.alsina@...>; Azmy Ariff <

azmyarif@...>; Chris Caldwell <

caldwell@...>; Pierre Deligne <

deligne@...>; Gaussianos <

gaussianos@...>; Andrew Granville <

andrew@...>; Lista NMBRTHRY <

nmbrthry@...>; Antonio Pérez Sanz <

aperez4@...>;

primenumbers@yahoogroups.com; Carlos Rivera <

crivera@...>; CarlosB Rivera <

cbrfgm@...>

Sent: Saturday, June 13, 2009 12:05:30 AM

Subject: [PrimeNumbers] Prime Numbers Equation

Hello:

If Goldbach Conjecture is True then:

For all p prime number p>=7 exists q and r also primes such that:

p**2+4(q+r)* *2=4p(q+r) +1

(It is easy to prove)

Sincerely

Sebastian Martin Ruiz

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed]