[seqfan] Re: Another duplicate: primes of the form x^2 + 4xy - 4y^2

[Alonso Del Arte]

Dear Vladimir:

Thank you very much. I'm starting to work on editing A141174 accordingly.

Al

On Thu, Jan 26, 2017 at 5:08 PM, Vladimir Shevelev <shevelev at bgu.ac.il>
wrote:

> Dear Al,
>
> A necessary and sufficient condition of representation of p=8n+1
> in your quadratic form is
> {8y^2+8n+1 is perfect square}, since only in this case
> solving square equation for x, we have
> x=-2y+sqrt(8y^2+8n+1) is integer.
> For this  a sufficient condition is { n has a form
> n=k^2-k + i(4k+i-1)/2, i>=0, k>=1}.
>  In this case  x=2i + 2k-1. y=k.
>
> Best regards,
> Vladimir
>
>
>
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Alonso Del Arte
> [alonso.delarte at gmail.com]
> Sent: 25 January 2017 19:15
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Another duplicate: primes of the form x^2 + 4xy - 4y^2
>
> I am much more confident about this other duplicate, but I still want to
> run it by other people before going ahead with it.
>
> A141174, primes of the form x^2 + 4xy - 4y^2, is a duplicate of A007519,
> primes of the form 8n + 1.
>
> I've already done the easy step, proving that all primes of the form x^2 +
> 4xy - 4y^2 are congruent to 1 mod 8. Since x^2 + 4xy - 4y^2 = 2 or -2 is
> impossible, x must be odd. And since x is odd, x^2 = 1 mod 8.
>
> If y is even, then both 4xy and 4y^2 are multiples of 8. If y is odd, then
> 4xy = 4 mod 8, but so is 4y^2, cancelling out the effect and leaving x^2 =
> 1 mod 8.
>
> There's still the issue of proving every prime of the form 8n + 1 has an
> x^2 + 4xy - 4y^2 representation. With the previous duplicate, this was
> proven with quadratic forms, if I recall correctly.
>
> Al
>
> --
> Alonso del Arte
> Author at SmashWords.com
> <https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>
>
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-- 
Alonso del Arte
Author at SmashWords.com
<https://www.smashwords.com/profile/view/AlonsoDelarte>
Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>