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

[Charles Greathouse]

Bravo!

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Thu, Jan 9, 2014 at 1:49 PM, David Applegate <david at research.att.com>wrote:

> 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
>
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>
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