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

Thu Jan 26 23:08:24 CET 2017

```Dear Al,

A necessary and sufficient condition of representation of p=8n+1
{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,

________________________________________
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

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

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