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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.

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    $\begingroup$ Set $y=z+\frac{1}{2}$, then the equation is $x^3=z^2+\frac{3}{4}$. $\endgroup$ – Dietrich Burde Mar 31 '15 at 7:48
  • $\begingroup$ See this and this. $\endgroup$ – Bumblebee Mar 31 '15 at 7:50
  • $\begingroup$ @Nilan: this does not look like an elliptic curve to me. $\endgroup$ – Luigi D. Mar 31 '15 at 8:23
  • $\begingroup$ 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. $\endgroup$ – DanielV Mar 31 '15 at 8:25
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    $\begingroup$ The first lines of mathworld.wolfram.com/MordellCurve.html may be inspiring. $\endgroup$ – Jack D'Aurizio Mar 31 '15 at 9:54
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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.

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  • $\begingroup$ 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. $\endgroup$ – Mirko Apr 17 '15 at 17:14

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