## Re: Diophantic equation and twin primes

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• ... Based on Warren s (corrected) simplification to your equation -3x^2 + y^2 -2xy - 4cx + 4cy + 4 = 0 namely r^2 - x^2 - c^2 + 1 = 0 then I can prove
Message 1 of 5 , May 26, 2013
--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
>
>
> Prove:
>
> Theorem:
>
> Let c a positive even number >2
>
> If x=1 and y=1 is the only positive solution of the polynomial
>
> -3x^2+y^2-2xy-4cx+4cy+4=0
>
>  then c +1 and c-1 are twin primes
>
> Sincerely
>
> Sebastián Martín Ruiz
>
> [Non-text portions of this message have been removed]
>

Based on Warren's (corrected) simplification to your equation

-3x^2 + y^2 -2xy - 4cx + 4cy + 4 = 0

namely

r^2 - x^2 - c^2 + 1 = 0

then I can prove something close, that if

(x,y) = (1,1) and ( (c^2)/2 - 1,3(c^2)/2 - 2c - 1)

are the only positive solutions to your equation, then c+1 and c-1 are twin primes.

If c is even and greater than 2, there are at least two unique solutions, as per above.

If c is odd and greater than 5, there are at least three unique solutions::

(x,y) = (1,1) and ((c^2 - 1)/4 - 1, 3(c^2 - 1)/4 - 2c + 1) and
((c^2 - 1)/8 - 2, 3(c^2 - 1)/8 - 2c + 2)

Mark
• Sorry If (1,1) and (c^2/2-1, 3c^2/2-2c-1) are the only integer solutions then c+1 y c-1 are twin primes Enviado desde Yahoo! Mail con Android [Non-text
Message 2 of 5 , May 26, 2013
Sorry If (1,1) and (c^2/2-1, 3c^2/2-2c-1) are the only integer solutions then c+1 y c-1 are twin primes

Enviado desde Yahoo! Mail con Android

[Non-text portions of this message have been removed]
• In http://tech.groups.yahoo.com/group/primenumbers/message/24217 SMR wrote: Sorry. I promise not to send anything if I m not sure that s interesting. ...
Message 3 of 5 , May 27, 2013
In
SMR wrote:
"
Sorry. I promise not to send anything if I'm not sure that's interesting.
"
when I said:
> It's not nice to obfuscate simple expressions
> to make serious people loose their time.

This was on 17.4.2012.
When I search "obfuscated Ruiz" in my mailbox,
the next earlier instance is from 16.3.2011,
and even earlier ones on 22.9.2010 and 31.8.2010.
Probably there are more using other terms....

Given that the case seems hopeless,
I finally didn't work out and send off the answer I was tempted to make.
M.

On Sun, May 26, 2013 at 6:38 PM, Sebastian Martin Ruiz <s_m_ruiz@...>wrote:

> **
> Sorry If (1,1) and (c^2/2-1, 3c^2/2-2c-1) are the only integer solutions
>

of the equation
A*B = (c-1)(c+1) ,
(where the above (x,y) correspond to A=1 resp. A=c-1)

> then c+1 y c-1 are twin primes
>

Obviously, indeed.
As usual, sufficiently obfuscated to have several people loose some hours
on that....

Maximilian

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