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

[David Wilson]

----- Original Message ----- 
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 6:57 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?

Yes, as in

1729 = 1^3 + 12^3 = 9^3 + 10^3

or the 3-way sum
15170835645 = 517^3 + 2468^3 = 709^3 + 2456^3 = 1733^3 + 2152^3