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 $p,q,m$?
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5$\begingroup$ What's the Motivation? $\endgroup$– Fan ZhengCommented Jan 11, 2017 at 1:01
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$\begingroup$ Might be useful: take the product of the two equations with one of them switching sides, divide both sides by $m^4$, then re-express both sides in terms of $\frac{p}{m}$ and $\frac{q}{m}$. You get that two quartics on different variables are equal. $\endgroup$– user44191Commented Jan 11, 2017 at 2:03
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1 Answer
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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.
answered Jan 11, 2017 at 1:09
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$\begingroup$ The genus $10$ curve $C$ has two automorphisms (which are more clear from the projective form of the original equation). The quotient curve $X$ has genus $3$ and three obvious rational points. $\endgroup$ Commented Jan 11, 2017 at 13:24