[seqfan] Re: Minimal k > n such that (4k+3n)(4n+3k) is square

[Robert G. Wilson v]

It was the discussion of this which led me to submit
https://oeis.org/draft/A235188 which is still not approved. Bob.

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of David
Applegate
Sent: Thursday, January 09, 2014 1:50 PM
To: seqfan at list.seqfan.eu
Subject: [seqfan] Re: Minimal k > n such that (4k+3n)(4n+3k) is square

Charles,
Thank you for this intriguing question.  I learned a lot thinking about it.
Here is a brief proof sketch (slightly more detailed than the comment I've
added to the OEIS entry for A083752):

If we let j := 24k + 25n, then (4k+3n)(4n+3k)=x^2 becomes
j^2 - 48 x^2 = 49 n^2.  For a fixed n, this is a Pell-like equation.
(Neil Sloane kindly pointed me to "Topics in Number Theory" by Leveque for
more information about Pell equations).  The corresponding Pell equation,
j^2 - 48 x^2 = 1, has a fundamental solution of j=7,x=1.

This means that if (j,x) is an integer solution to j^2-48x^2=49n^2, then so
are (7j+48x,7x+j) and (7j-48x,7x-j).

For since k = (j-25n)/24, for j to correspond to an integer k, we must have
j == n (mod 24).  If j satisfies this, though, 7j+48x will not.
However, if we iterate again, we get that if (j,x) is an integer solution to
j^2-48x^2=49n^2 with j == n (mod 24), then so are
(97j+672x,97x+14j) and (97j-672x,97x-14j).

Finally, if we iterate this again, and translate it back in terms of k,
after a bunch of manipulation we get that if (k,x) is an integer solution to
(4k+3n)(4n+3k)=x^2, then so is (18817k+19600n-5432x,18817x-65184k-67900n).

If we define r := k/n (or, equivalently, k = r n), then x = sqrt(12 k^2 + 25
kn + 12 n^2) = n sqrt(12 r^2 + 25 r + 12), and we have 18817k + 19600n -
5432x = n (18817 r + 19600 - 5432 sqrt(12 r^2 + 25 r + 12))

Put another way, this means that if we have a solution with k/n = r, then
there is also a solution with

k/n = f(r) = (18817 r + 19600 - 5432 sqrt(12 r^2 + 25 r + 12))

f(r) is strictly decreasing on the interval [1,393], with f(1)=393,
f(109/4)=109/4, and f(393)=1.

In paticular, then, this means that if we have a solution k with
109/4 < k/n < 393, then there is also a solution k1 with k1/n = f(k/n), so 1
< f(k/n) < 109/4.

-Dave

> From seqfan-bounces at list.seqfan.eu Fri Dec 20 16:30:38 2013
> From: Charles Greathouse <charles.greathouse at case.edu>
> Date: Fri, 20 Dec 2013 16:29:48 -0500
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Minimal k > n such that (4k+3n)(4n+3k) is square

> I noticed that a(n) < 393*a(n) for
> http://oeis.org/A083752
> and wondered if this could be sharpened. It seems that for many n, 
> a(n) = 393n, but if not then a(n) <= 27.25n, with many n (obviously 
> multiples of
> 4) having this precise value.

> The graph of A083752/A27 is compelling. The following conjecture is 
> numerically natural, though I have no theoretical reason to believe it
yet:

> If there is no n < k < 109n/4 with (4k+3n)(4n+3k) square, then a(n) =
393n.

> Can someone find either a counterexample or a proof?

> I have a feeling that the method of a proof might lead to a more 
> efficient algorithm for the sequence (aside from the obvious 
> optimization of skipping the associated k).

> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University

> _______________________________________________

> Seqfan Mailing list - http://list.seqfan.eu/


_______________________________________________

Seqfan Mailing list - http://list.seqfan.eu/