[seqfan] Re: a^2 + b^3 = c^4

Alonso Del Arte alonso.delarte at gmail.com
Mon May 5 16:08:07 CEST 2014


I would add the "c" sequence first, and hold off on "a" and "b" until I or
someone else can resolve the theoretical questions, like which which values
of c have more than one pair of a and b, and whether there is such a thing
as a "primitive" solution.

Al


On Mon, May 5, 2014 at 3:00 AM, Lars Blomberg <lars.blomberg at visit.se>wrote:

> Putting each of a,b,c in increasing order is the logical thing to do, I
> agree.
>
> I have computed some c values and my (somewhat hasty) thought was to
> include the
> corresponding a and b values as separate sequences.
> But as you point out, this will not be correct.
>
> Maybe I will stick with the "c" sequence for the time being.
>
> /Lars
>
> -----Ursprungligt meddelande----- From: israel at math.ubc.ca
> Sent: Monday, May 05, 2014 8:35 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: a^2 + b^3 = c^4
>
>
> What order would you put these in?
> It would seem logical to put them each in increasing order.
> Thus the "a" sequence would be the set of all a such that
> a^2 + b^3 = c^4 for some b and c.
> However, searching for solutions may be difficult: I don't know
> if there are effective bounds on b and c for given a.
> The other two should be OK: for the "c" sequence we certainly have
> a < c^2 and b < c^(4/3), while for the "b" sequence, since
> b^3 = (c^2+a)(c^2-a) > c^2 + a, so c < b^(3/2) and a < b^3.
>
> Robert Israel
> University of British Columbia and D-Wave Systems
>
> On May 4 2014, Lars Blomberg wrote:
>
>  Hello Seqfans,
>>
>> a^2 + b^3 = c^4 has solutions
>> a = 28, 27, 63, 1176, 648, 433, 1792, ...
>> b = 8, 18, 36, 49, 108, 143, 128, ...
>> c = 6, 9, 15, 35, 36, 42, 48, ...
>> none of which seem to be in OEIS.
>>
>> I intend to add the "c" sequence.
>> One question though: Should I add the "a" and "b" sequences as well?
>>
>> /Lars
>>
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>>
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>>
>>
>>
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Alonso del Arte
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