## Re: [PrimeNumbers] Prime Numbers Equation

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• 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
Message 1 of 7 , Jun 12, 2009
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

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

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

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