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

[David Wilson]

Never mind.

My brain is totally rotted now.

I knew I should stay out of the discussion.

----- Original Message ----- 
From: "David Wilson" <davidwwilson at comcast.net>
To: "Hans Havermann" <pxp at rogers.com>; "James Buddenhagen" 
<jbuddenh at earthlink.net>
Cc: <seqfan at ext.jussieu.fr>
Sent: Thursday, October 28, 2004 11:54 AM
Subject: Re: Divisible by 12: a^3 + b^3 = c^2


>
> ----- 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
>
>
>