Fw: a Diophantine equation question
Jon Schoenfield
jonscho at hiwaay.net
Mon May 28 21:00:16 CEST 2007
All,
Jaap Spies brought to my attention that I had given an incorrect expression
for the sum of cubes from (c-k)^3 to (c+k)^3; the result should have been
(c-k)^3 + ... + (c+k)^3 = ... = (2k+1)c^3 + k(k+1)(2k+1)c
('Sorry about that!) :-/
So, for instance, at the value of k in Jaap's email to me below, i.e., k =
31, the sum of 63 consecutive cubes with c^3 as the central value is
(c-31)^3 + ... + (c+31)^3 = ... = 63c^3 + 62496c
and the values of c at which the sum is a square are the usual ones, i.e.,
c = 0
c = k = 31
c = k+1 = 32
c = 2k(2k+1)(2k+2) = 249984
and also the (unusual) values
c = 4
c = 248 (which is 31 * 8)
c = 868 (which is 31 * 28)
c = 1152 (which is 32 * 36)
... and maybe others ...?
(Thanks again, Jaap!)
-- Jon
----- Original Message -----
From: "Jaap Spies" <j.spies at hccnet.nl>
To: <jonscho at hiwaay.net>
Cc: <njas at research.att.com>
Sent: Monday, May 28, 2007 6:36 AM
Subject: Re: a Diophantine equation question
> Hi Jon,
>
> You wrote:
>
>>>
>>> Having run similar tests for squares that are equal to the sum of 2k+1
>>> consecutive cubes (for positive integer values of k), and noting that
>>>
>>> (c-k)^3 + ... + (c+k)^3 = ... = (2k+1)c(c^2 + 2k(2k+1)(2k+2))
>>>
>
> This formula seems to be incorrect:
>
> I think we have s^2 = (2k+1)c^3 + k(k+1)(2k+1)c, substituting X=(2k+1)c,
> Y=(2k+1)s
> we get the elliptic curve
>
> E_k: Y^2 = X^3 + k(k+1)(2k+1)^2 X
>
>>> it looks to me as though, for _most_ values of k, there will exist only
>>> four values in the sequence (as I believe is the case for A116108): the
>>> ones corresponding to c = 0, k, k+1, and 2k(2k+1)(2k+2). The only
>>> exceptions I've found (through k = 50) are as follows (and if there's
>>> any kind of pattern here, I haven't yet figured it out):
>>>
>
> The number of integral solutions can not be predicted as it depends on
> properties of the elliptic curve at hand. For k=31 E is of rank 3, there
> are 3 independent generators and quite a few integral solutions.
>
> Regards,
>
> Jaap
>
>
>
>
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