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

[James Buddenhagen]

From: "Hans Havermann" <pxp at rogers.com>
To: "James Buddenhagen" <jbuddenh at earthlink.net>
Cc: <seqfan at ext.jussieu.fr>
Sent: Wednesday, October 27, 2004 5:57 PM
Subject: Re: Divisible by 12: a^3 + b^3 = c^2


> James Buddenhagen wrote:
> 
> > Lemma 2:  If gcd(a,b)=1 and a^3 + b^3 = c^2 then either
> >   (1)  a+b and a^2-a*b+b^2 are both squares or
> >   (2)  a+b and a^2-a*b+b^2 are both 3 times squares.
> 
> 
> Is it possible for c^2 to be a "taxicab" number, c^2 = a1^3 + b1^3 = 
> a2^3 + b2^3, where both gcd(a1,b1)=1 and gcd(a2,b2)=1?

I don't know, but if you don't require gcd=1 then one finds

2457^3 + 3276^3 = 897^3+3666^3 = 223587^2
1026^3+1710^3 = 228^3+1824^3 = 77976^2

Maybe someone else can find with gcd=1.

Jim Buddenhagen