Divisible by 12: a^3 + b^3 = c^2

[James Buddenhagen]

----- Original Message ----- 
From: "Ralf Stephan" <ralf at ark.in-berlin.de>
To: "Pfoertner, Hugo" <Hugo.Pfoertner at muc.mtu.de>
Cc: "'Ed Pegg Jr'" <edp at wolfram.com>; "seqfan" <seqfan at ext.jussieu.fr>
Sent: Thursday, October 21, 2004 4:05 AM
Subject: Re: Divisible by 12: a^3 + b^3 = c^2


> There is no c/12 == 3 mod 5 in the list.
> Added c/12 mod 7 column without further comment.
> Also the * are c/12 == 17 mod 19.
> 
> >       a       b          c       c/12  (c/12) % 7
> > 
> >      11      37        228         19           5
> >    2137    8663     812340      67695           5  *
> >    8929   28703    4935504     411292           0
> >    1801   56999   13608420    1134035           0
> >   44111   48817   14218512    1184876           0  *
> >    6637   86291   25354032    2112836           5  *
> >   57241   87959   29463060    2455255           5
> >   16931  133597   48880608    4073384           0
> >  151477  246011  135516108   11293009           0  *
> >   54083  334717  194057640   16171470           0
> >    3889  367823  223078944   18589912           5
> >  127717  552011  412662012   34388501           0
> >  457511  786697  763311948   63609329           0
> >  571019  735781  764539380   63711615           2  *
> >  765983 1154017 1409359200  117446600           5
> >   38557 1586531 1998370272  166530856           2  *
> 
> ralf

Here are some observations:

Let a = 30000*m^4+72600*m^2-14641, b = -30000*m^4+72600*m^2+14641 
and c = 660*m*(30000*m^4+14641).  Then a^3 + b^3 = c^2 identically 
and specializing to m = 1 gives [a,b,c] = [87959, 57241, 29463060], 
which corresponds to the row of your table which I used to find the 
identity (I believe a,b are reversed from what you had).

Similarly, each row of your table gives rise to an identity.
Quite often a,b are both prime or negatives of primes for 
integer m.  To get a, b both positive we may use rational m and 
adjust to find integer solutions to a^3 + b^3 = c^2. 

Searching on the family above I quickly found over 600 solutions 
with a, b both prime.  Here are the first few

       a               b                 c/12     @2   @3   @5   @7  @11  @13  @17  @19

     44111           48817              1184876    0    2    1    0    0    4   10   17
     87959           57241              2455255    1    1    0    5    0   10   13   18
    367823            3889             18589912    0    1    2    5    0    3    4    8
    571019          735781             63711615    1    0    0    2    0    6    1   17
   4683779         9836221           2705984050    0    1    0    0    0    3    6    3
   8142803        11395309           3745021082    0    2    2    5    0   10   13   14
  10185443        12140509           4445725438    0    1    3    0    0   11    1    7
  23313383        29103817          16099304485    1    1    0    5    0    4   15    0
  41436323        14343709          22683813922    0    1    2    0    0    3    7    5
   9979511        72267577          51263122141    1    1    1    0    0   10    0    1
  22471223        80266489          60580560887    1    2    2    0    0    2    2    0
   1450739       112386061          99285972940    0    1    0    0    0   11    7    3
  26613731       129598237         123478061476    0    1    1    0    0    3   16   16
 165560543        22618657         177748942140    0    0    0    0    0    3   15    1
  81178931       160526797         180114971916    0    0    1    5    0    9    0    0
 229884143       138947089         320927216148    0    0    3    0    0    6    2    1
 249609911       139996537         356446278881    1    2    1    0    0    4    8    8
 315400559       139946641         486741766280    0    2    0    0    0    3   12   18
 165272123       375017077         630563939665    1    1    0    2    0   10   15   16
  24006119       415223881         705154365025    1    1    0    5    0    6   15   17
 255985151       556414657        1145761379026    0    1    1    5    0    4    0    1
 288116519       572774281        1212856911035    1    2    0    0    0    3    9    0
 620586959        90893041        1290339207850    0    1    0    0    0    4    2   16
 726190931       221767597        1653837884116    0    1    1    5    0    2   10    0
 554843903       684838849        1848426508698    0    0    3    0    0    3   15   18

The columns headed with @2, @3, @5 etc give c/12 mod 2, c/12 mod 3, c/12 mod 5 etc.  
Note that c/12 = 3 mod 5 occurs several times.  

Here a = -1 mod 4 and b = 1 mod 4.  Also, a^3 + b^3 = (a+b)*(a^2-a*b+b^2) and each 
of a+b and a^2-a*b+b^2 equals 3 times a square.

I suspect that there is an easy characterization of which a,b give c divisible by 12,
but that is all I did with this for now.

Jim Buddenhagen