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

Robert G. Wilson v rgwv at rgwv.com
Mon May 5 15:50:31 CEST 2014

```SeqFans,

Sorted, I obtain the following terms:

for a: 27, 28, 63, 433, 648, 1176, 1728, 1792, 2925, 3807, 4032, 4500, 4785, 4941, 6000, 6083, 6875, 7203, 7452, 7902, 8100, 10000, 10125, 12005, 13328, 14703, 15525, 19683, 20412, 21266, 26775, 27712, 32507, 33750, 35672, 40572, 40797, 41328, 41472, 45927, 49375, 50625, 51313, 55625, 67400, 73205, 73695, 75264, 83511, 83853, 85775, 88125, 95992, ..., .
for b: 8, 18, 23, 36, 49, 108, 108, 126, 128, 135, 136, 143, 216, 225, 245, 288, 288, 300, 343, 368, 375, 400, 450, 500, 576, 588, 600, 648, 686, 693, 784, 900, 1026, 1125, 1156, 1183, 1215, 1350, 1350, 1458, 1568, 1628, 1638, 1728, 1728, 2000, 2016, 2028, 2048, 2160, 2176, 2178, 2178, 2288, 2646, 2790, 2916, 3375, 3456, 3528, 3600, 3600, 3969, 4225, 4375, 4590, 4608, 4913, 5000, 5046, 5400, 5415, 5488, 5808, 6000, ..., .
for c: 6, 9, 15, 35, 36, 42, 48, 57, 63, 71, 72, 75, 78, 90, 98, 100, 120, 135, 141, 147, 162, 195, 196, 204, 208, 215, 225, 243, 252, 260, 279, 280, 288, 289, 295, 300, 306, 336, 363, 364, 384, 405, 441, 450, 456, 462, 504, 510, 525, 537, 550, 568, 576, 600, 624, 630, 713, 720, 722, 735, 750, 784, 800, 819, 828, 841, 845, 847, 867, 875, 891, 909, 925, 945, 946, 960, 972, 980, ..., .

Several c's have more than one solution, example: c=72; 72^4 = 1728^2 + 288^3 = 4941^2 + 135^3.
The list of c's having more than one solution is: {72, 100, 147, 225, 456, 576, 800}

I am not confident that I have found all a's or b's without going further with the c's.

RGWv

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Lars Blomberg
Sent: Monday, May 05, 2014 3:00 AM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: a^2 + b^3 = c^4

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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>
>Seqfan Mailing list - http://list.seqfan.eu/
>
>

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