[seqfan] Re: a^2 + b^3 = c^4
Lars Blomberg
lars.blomberg at visit.se
Mon May 5 09:00:10 CEST 2014
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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