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

[Dean Hickerson]

Ed Pegg Jr wrote:

> 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:
...
> Seems like it should be extendable and provable.

(I'm changing a and b to p and q, since that's what I have in my notes,
and if I try to change them all I'll probably miss some.)

Here's a proof.  First, if p=2, then  q^3 + 8 = c^2.  According to

    http://mathworld.wolfram.com/MordellCurve.html

this equation is known to have only 4 solutions in integers, namely
(q,c) = (-2,0), (1,3), (2,4), (46,312).  None of these have q prime and
p<q, so from now on we may assume that both p and q are odd primes.

Factoring, we have

    (p+q) (p^2 - p q + q^2) = c^2.                                      (0)

Let

    d = gcd(p+q, p^2 - p q + q^2).                                      (1)

Then d divides

    (p^2 - p q + q^2) + (p+q) (2p - q) = 3 p^2;

similarly d divides  3 q^2.  Since p and q are distinct primes, d = 1 or 3.

Suppose that d=1.  Then both  p+q  and  p^2 - p q + q^2  are squares.
Let

    p^2 - p q + q^2 = r^2                                               (2)

with r positive.  Then

    p q = r^2 - (p-q)^2 = (r+p-q) (r+q-p).                              (3)

Since  p < q  and  0 < r+p-q < r+q-p,  we have either

    r+p-q = 1  and  r+q-p = p q                                         (4)

or

    r+p-q = p  and  r+q-p = q.                                          (5)

(4) implies  p q - 1 = 2q - 2p,  so  (p-2)(q+2) = -3,  contradicting the
fact that p and q are >= 3.  (5) implies  r = p = q,  contradicting  p<q.

Hence d=3, so both  p+q  and  p^2 - p q + q^2  are squares multiplied by 3.
Let

    p^2 - p q + q^2 = 3 s^2.                                            (6)

Since p and q are odd, so is s; hence

    p^2 == q^2 == s^2 == 1 (mod 8),                                     (7)

which implies

    p q == 7 (mod 8).                                                   (8)

Therefore 8 divides  p^2 + p q = p (p+q),  so 8 divides p+q.  Finally,
we have

    c^2 = (p+q) (p^2 - p q + q^2) = (p+q) 3 s^2,                        (9)

so c^2 is divisible by 24 and c is divisible by 12.

Dean Hickerson
dean at math.ucdavis.edu