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?