[seqfan] Re: a^2 + b^3 = c^4
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Tue May 6 12:18:48 CEST 2014
I'm not sure how the c sequence works as an index on the other two, since for the same
value of c there could be more than one pair (a,b) which works.
For fixed c, this is an elliptic curve in the (a,b) plane. It is more commonly written as
y^2 = x^3 + d
taking (x,y,d) = (-b, a, c^4). Siegel's theorem says that for fixed d (i.e. fixed c) there are
only finitely many integer solutions.
For a beautiful introduction to Siegel's theorem see the article
http://arxiv.org/pdf/1005.0315v3.pdf
which is published in American Mathematical Monthly.
All the best
Andy
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Charles Greathouse [charles.greathouse at case.edu]
Sent: 05 May 2014 15:50
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: a^2 + b^3 = c^4
That's the way I would do it -- add c, then a and b sequences indexed on c.
Charles Greathouse
Analyst/Programmer
Case Western Reserve University
On Mon, May 5, 2014 at 10:38 AM, Jean-François Alcover <jf.alcover at gmail.com
> wrote:
> I agree: the c-sequence should be the first, and, in my opinion,
> if should serve as index to the a- and b-sequence,
> and show the duplicates this way:
> 6, 9, 15, 35, 36, 42, 48, 57, 63, 71, 72, 72, 75, 78, 90, 98, 100,
> 100, 120, 135, 141, 147, 147, 162, 195, 196, 204, 208, 215,
> 225, 225, 225, 243, ...
>
> jfa
>
> 2014-05-05 16:08 GMT+02:00 Alonso Del Arte <alonso.delarte at gmail.com>:
>
> > 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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> >
> >
> > --
> > Alonso del Arte
> > Author at SmashWords.com<
> > https://www.smashwords.com/profile/view/AlonsoDelarte>
> > Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>
> >
> > _______________________________________________
> >
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> >
>
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