[seqfan] Re: A236767

[David Applegate]

Yes.  If x^2 = y^4 + p, let a = x - y^2.  Then

y^4 + p = x^2 = (y^2 + a)^2 = y^4 + 2 a y^2 + a^2

so

p = 2 a y^2 + a^2, and so a divides p.  Since p is a prime, a must be
a unit (that is, +1 or -1).

But since p >= 2, a must be +1.

-Dave

> From seqfan-bounces at list.seqfan.eu Fri Jan 31 10:48:46 2014
> From: Hans Havermann <gladhobo at teksavvy.com>
> Date: Fri, 31 Jan 2014 10:48:28 -0500
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] A236767

> I'm currently contributing/editing this new sequence: 2, 10, 37, 82, 442, 577, ..., numbers whose square is a fourth power plus a prime. After a day of working on a b-file, I noticed that - up to the number of terms that I had calculated (just short of 500) - this sequence matched A089001^2 + 1. It's easy to show that A089001^2 + 1 will always be a member but are they the *only* members?

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