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

[Ed Pegg Jr]

Kirk Bresniker notes that for a^3 + b^3 = c^2, if a and b
are prime, then for a<b<100000, c has a factor of 12:

11^3    + 37^3    = (2 * 2 * 3  * 19)^2
1801^3  + 56999^3 = (2 * 2 * 3  * 5 * 7  * 32401)^2
2137^3  + 8663^3  = (2 * 2 * 3  * 3 * 5  * 4513)^2
6637^3  + 86291^3 = (2 * 2 * 3  * 2 * 2  * 11 * 31 * 1549)^2
8929^3  + 28703^3 = (2 * 2 * 3  * 2 * 2  * 7  * 37 * 397)^2
44111^3 + 48817^3 = (2 * 2 * 3  * 2 * 2  * 7  * 11 * 3847)^2
57241^3 + 87959^3 = (2 * 2 * 3  * 5 * 11 * 44641)^2

Seems like it should be extendable and provable.

--Ed Pegg Jr