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


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.,


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.

  • 1
    $\begingroup$ 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

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);

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.

  • $\begingroup$ 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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