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

[Paul C. Leopardi]

Hi all,
We have
a^3 + b^3 = (a+b)^3 - 3*a^2*b - 3*a*b^2
          = (a+b)^3 - 3*a*b*(a+b)
          = (a+b)*((a+b)^2 - 3*a*b).
 From this, it is fairly easy to prove that 
if a^3 + b^3 = c^2, then
1) If a and b are odd, then a+b is divisible by 4 and c is divisible by 2.
2) If, additionally, a+b is divisible by 3, then c is divisible by 6.

Note that in all cases originally posted by Ed Pegg Jr, 
we have a+b = 48*d^2, where

    a     b  d         a+b = 48*d^2
   11    37  1            48
 1801 56999  5 * 7     58800
 2137  8663  3 * 5     10800
 6637 86291  2^2 * 11  92928
 8929 28703  2^2 * 7   37632
44111 48817  2^2 * 11  92928
57241 87959  5 * 11   145200 

We also have in all cases above,
a+b = 48*e, where e = ((2^f)*(3^g))^2 * h, 
with f and g non-negative integers and with h not divisible by 2 or 3.

It is fairly easy to prove that:

Lemma
If a and b are odd primes, a^3+b^3=c^2,
and a+b = 48*e, where e = ((2^f)*(3^g))^2 * h, with f and g non-negative 
integers and with h not divisible by 2 or 3, 
then c is divisible by 12.

Proof:
c^2 = 48*(((2^f)*(3^g))^2)*h*((48^2)*((((2^f)*(3^g))^2)*h)^2 - 3*a*b)
    = (12^2)*(((2^f)*(3^g))^2)*h*(3*(16*((((2^f)*(3^g))^2)*h)^2 - a*b).

If we let i := 3*(16*((((2^f)*(3^g))^2)*h)^2 - a*b,
then we see that i is not divisible by 2 or 3.
Then h*i is not divisible by 2 or 3, so h*i must be a square, h*i=j^2,
with j not divisible by 2 or 3.
So then
c = 12 * 2^f * 3^g * j.
[]

That's as far as my analysis has gone (modulo typos). I can give more details, 
if anyone is interested.
Best regards

On Friday 22 October 2004 08:09, James Buddenhagen wrote:
> ----- 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
[...]
> >
> > 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
[...]
> 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