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## Diophantine equation and twin primes

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• ________________________________ Prove:  Theorem: Let c a positive even number 2 If (x,y)= (1,1) and (c^2/2-1,3c^2/2-2c-1) are the only positive integer
Message 1 of 2 , May 26, 2013
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Prove:

Theorem:

Let c a positive even number >2

If (x,y)= (1,1) and (c^2/2-1,3c^2/2-2c-1) are the only positive integer solutions

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 (r-x)(r+x) =
Message 2 of 2 , May 27, 2013
--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
>
> ________________________________
>
> Prove:
>
> Theorem:
>
> Let c a positive even number >2
>
> If (x,y)= (1,1) and (c^2/2-1,3c^2/2-2c-1) are the only positive integer solutions
>
>  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
>

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

(r-x)(r+x) = (c-1)(c+1)

If both c-1 and c+1 are prime, then clearly there are only two options:

r-x = c-1 and r+x = c+1

OR

r-x = 1 and r+x = (c-1)(c+1)

Follow each of those options and you easily arrive at the two results.

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

Mark
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