On $x^3-y^2=1728 \text{ unit}$ in number fields

Consider solution of $$x^3-y^2=1728 \text{ unit} \qquad (1)$$ in a number field.

This is related to the discriminant of elliptic curve in terms of $c_4,c_6$.

Via elliptic curves it might have infinitely solutions for fixed unit.

I am interested if $x,y$ are in the ring of integers of the number field.

If one solution exists, scaling by powers of units gives infinitely, so define a set of solutions to be primitive if the norms of $x$ are distinct.

Q1 Does (1) have infinitely many primitively solutions when $x,y$ are in the ring of integers?

Q2 Does positive solution to Q1 contradicts the abc conjecture over number fields?

Q3 What is the expected abc quality of triples resulting from (1)? Say for $w=\sqrt{2}$, A= (w + 1) * 3^3 ;B= (-70*w - 99) * w; C= (-2*w - 1)^4, A+B+C=0

I suppose this is impossible since it might solve $x^3-y^2=u^n$ for many $n$.

• The equation $y^2 = x^3 -1726\,u$ for a fixed unit $u$ is an elliptic curve and so it has finitely many integral points by Siegel's theorem. So I read the question as asking to find solutions when $u$ is allowed to vary in all units. – Chris Wuthrich Mar 14 '14 at 17:23
• @ChrisWuthrich indeed, $u$ varies in infinitely many units. – joro Mar 14 '14 at 17:46

You can always absorb 6th powers into $x$ and $y$, so there is a finite set of units $u_1,\ldots,u_n$ (covering the congruence classes of units mod 6th powers), so that every solution to $x^3-y^2=1728u$ with $x,y\in\mathcal{O}_K$ and $u\in\mathcal{O}_K^*$ yields a solution $(X,Y)\in\mathcal{O}_K\times\mathcal{O}_K$ to one of the finitely many equations $x^3-y^2=1728u_i$ with $1\le i\le n$. Each of these finitely many equations has finitely many solutions (by Siegel). From this it is easy to answer your question, Start from the finitely many solutions $(X_j,Y_j)$ with $1\le j\le N$, then the full set of solutions to your equation is $$\{ (v^2X_j,v^3Y_j) : 1\le j\le N\quad\hbox{and}\quad v\in\mathcal{O}_K^*\}.$$ Since the units have norm 1, there are only finitely many solutions with distinct $x$-norms.
• I still don't understand why the infinite set $u_i$ must have finitely many solutions with distinct norms. If you take finite sets $u_i$, the set of integral points is clearly finite. – joro Mar 14 '14 at 18:26
• $\mathcal{O}^*_K$ modulo $6$-th powers is a finite group. – Chris Wuthrich Mar 14 '14 at 23:20