[seqfan] Re: p = (k + 1)^2 - k = (m + 1)^3 - m and p = (k + 1)^2 + k = (m + 1)^2 + m

[Jack Brennen]

Basically, you're looking for integer solutions to the elliptic curve:

    y^2 = x^3 - 16*x + 16

For any such solution (x,y), then you could have a solution with
(k,m) = ((y-4)/8,x/4-1).  (Assuming k and m are integers.)

The only integer solutions I can find to the elliptic curve are these
six:  (-4,4), (0,4), (1,1), (4,4), (8,20), (24,116), which yield
five integer solutions for (k,m):

Three trivial solutions:  (k,m) = (0,-2), (0,-1), (0,0)
Two non-trivial solutions:  (k,m) = (2,1), (14,5)




On 9/20/2013 2:26 PM, Alonso Del Arte wrote:
> So you're looking for numbers that are both central polygonal numbers (
> A002061 <http://oeis.org/A002061> <http://oeis.org/A002378>) and close to a
> cube (A061600 <http://oeis.org/A061600>)?
>
> Al
>
>
> On Fri, Sep 20, 2013 at 1:53 PM, юрий герасимов <2stepan at rambler.ru> wrote:
>
>>
>> Dear SeqFans,
>> p = (k + 1)^2 - k = (m + 1)^3 - m:
>>
>> 7 = (2 + 1)^2 - 2 = (1 + 1)^3 - 1,
>>
>> 211 = (14 + 1)^2 -14 = (5 + 1)^3 - 5,
>> What is the next one?
>>
>> p = (k + 1)^2 + k = (m + 1)^3 + m:
>> 29 = (4 + 1)^2 + 4 = (2 + 1)^3 + 2,
>> What is the next one?
>> Best regards,
>> JSG
>>
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>
>
>