[seqfan] Re: A362564

[Max Alekseyev]

I'd like to mention that the problem of finding (largest) x satisfying n +
2^x = y^2 can be solved via reduction to three Mordell curves:
n + z^3 = y^2,
n + 2z^3 = y^2 or equivalently 4n + (2z)^3 = (2y)^2,
n + 4z^3 = y^2 or equivalently 16n + (4z)^3 = (4y)^2,
where z := 2^floor(x/3).
For a given n, each of these three curves is known to have only a finite
number of integer points (y,z), proving that x cannot be unbounded.

In many cases the actual points can be computed from scratch (e.g., in
Sage) or taken from the tables of known points on Mordell curves - see
External links section in https://en.wikipedia.org/wiki/Mordell_curve
Clearly, we'd be interested only in those points where z is a power of 2.

Regards,
Max


On Wed, Apr 26, 2023 at 7:40 AM Yifan Xie <xieyifan4013 at 163.com> wrote:

> Hi,
> Can anyone help me with the sequence A362564?
> Thomas Scheuerle and I are trying to prove a vaild upper bound for A363564
> or A238454 (see the discussion).
> Both of us have did a wrong proof...
>
>
> Best regards
> Yifan Xie (xieyifan4013 at 163.com)
>
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