Q:

How to solve the below simultaneouly Diophantine equations?

How to solve the simultaneous Diophantine equations:
$q=p^3+6m^2p-12m^3-12mp^2$,
$p=q^3+6m^2q-28m^3-28mq^2$
for rational numbers $p,q,m$?

A:

According to Maple, the resultant of the polynomials (left side - right side) with respect to $p$ is 
$$-21952\,{m}^{9}+14112\,{m}^{8}q-68880\,{m}^{7}{q}^{2}+30792\,{q}^{3}{m
}^{6}-69888\,{m}^{5}{q}^{4}+18924\,{m}^{4}{q}^{5}-23044\,{m}^{3}{q}^{6
}+2370\,{m}^{2}{q}^{7}-84\,m{q}^{8}+{q}^{9}-9408\,{m}^{7}+4032\,{m}^{6
}q-19248\,{m}^{5}{q}^{2}+4704\,{m}^{4}{q}^{3}-9552\,{m}^{3}{q}^{4}+672
\,{m}^{2}{q}^{5}-12\,m{q}^{6}-168\,{m}^{5}+36\,{m}^{4}q-168\,{m}^{3}{q
}^{2}+6\,{m}^{2}{q}^{3}-12\,{m}^{3}-q
$$
which is a curve of genus $10$.  By Faltings's theorem, there are only finitely many rational points.
Among the solutions are $(p,q,m) = (-1,-1,0), (0,0,0), (1,1,0)$.  There don't seem to be others with small numerators and denominators.

