Solve $x^3=y^2-y+1$ in positive integers.

Question

I recently started doing number theory and have finished with all the basic, intermediate and some of the advanced stuff with ease. However, I encountered this question and have been stuck for about a day with this

Solve in positive integers

$$x^3=y^2-y+1$$

I have tried modular arithematic and factorization but nothing seems to work so far. I've only been able to reduce it into an equivalent Diophantine Equation i.e.,

$$4x^3=y^2+3$$

Further, I'm not yet acquainted with algebraic, analytic or geometric number theory, so I'd prefer an elementary solution.

Any help will be appreciated.

Set $y=z+\frac{1}{2}$, then the equation is $x^3=z^2+\frac{3}{4}$. — Dietrich Burde, Mar 31 '15 at 07:48
Emperically, the only solutions less than 2 million are $(1,~0)$, $(1,~1)$, and $(7, 19)$, so proof techniques for showing only a finite number of solutions are probably preferred. @Luigi my first thoughts too, but I think he meant after the $y=z+1/2$ transform. — DanielV, Mar 31 '15 at 08:25
@LuigiD. It is an elliptic curve, since it is of the form $y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$ with nonzero discriminant. — Dietrich Burde, Mar 31 '15 at 09:43
The first lines of http://mathworld.wolfram.com/MordellCurve.html may be inspiring. — Jack D'Aurizio, Mar 31 '15 at 09:54
probably I have an argument to show the equation has finite solution, but can't provide solutions. — Michael, Apr 14 '15 at 06:32
@Jim Can you please tell me your reasoning. It'd be a great help. — , Apr 17 '15 at 11:22
An elementary start (which I do not know what to do with) would be to rewrite as $(x^2+x+1)(x-1)=y(y-1)$, at least shows you $x$ must be odd (which you could see anyway in the original form), so I do not know if this gives you something to work with. — Mirko, Apr 17 '15 at 15:48
The "natural" way of solving this would be algebraic number theory, via Eisenstein integers — Asier Calbet, Mar 19 '17 at 14:27

score 3 · Answer 1 · answered Apr 17 '15 at 15:03

This is supposed to be a comment but it is a little bit long.

The equation $y^2 - y + 1 = x^3$ can be rewritten as

$$Y^2 = X^3 - 48\quad\text{ where }\quad\begin{cases} Y &= 8y - 4\\ X &= 4x\end{cases}$$ If one throw following command to online Magma calculator

Q<x> := PolynomialRing(Rationals());
E00  := EllipticCurve(x^3-48);
Q00  := IntegralPoints(E00);
Q00;

Magma find only two pairs of integral solutions $(X,Y) = (4,\pm 4)$ and $(28,\pm 148)$. This corresponds to the solutions $(x,y) = (1,0 \verb/ or / 1 )$ and $(7, 19 \verb/ or / -\!\!18)$ It looks like these are the only integral solutions of the original equation.

I just connected you answer with the link in the comment by @JackD'Aurizio that Morldell's equation $Y^2=X^3+n$ has only finitely many solutions. — Mirko, Apr 17 '15 at 17:14

Solve $x^3=y^2-y+1$ in positive integers.

1 Answers1

Linked