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So, I'm looking for a rational solution of $$x^3-3x^2+3x+1=y^2$$ that is different from $(0,\pm 1)$. I tried to substitute $y=xt+1, t\in\mathbb Q$, getting following equation $$ x(x^2-(t^2+3)x+(-2t+3))=0$$ with the discriminant of the quadratic term $t^4+6t^2+8t-3$. The next step would be to find a $t$ for that this discriminant is a square of a rational number. Could you give me a tip how to do it or propose another solution?

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    $\begingroup$ Is there some particular reason for avoiding the simple equivalent forms $$x^3+2 = y^2$$ or $$(x+1)(x^2-x+1) = (y-1)(y+1) $$ ? $\endgroup$ Nov 3, 2016 at 21:10
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    $\begingroup$ The equation can be rewritten as $y^2=(x-1)^3+2$. $\endgroup$
    – Anurag A
    Nov 3, 2016 at 21:10
  • $\begingroup$ The question had a hint to consider the intersection of a secant at $(0,1)$ and the curve, so I tried to do it. $\endgroup$
    – user343447
    Nov 3, 2016 at 21:51

3 Answers 3

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Your equation rewrites as $(x-1)^3 + 2 = y^2$

Solve (or rather find a solution of) $x^3 + 2 = y^2$ for rationnals.

Write $x={p\over q}$ and $y={a\over b}$ where $a\land b =p\land q =1$.

You then have $\left({p\over q}\right)^3+2 = \left({a\over b}\right)^2$ and therefore $q^3 = b^2$. What follows is that there exists $k\in\Bbb N^*$ such that $q= k^2$ and $b=k^3$ (assuming $b\ge1$ and $q\ge1$). Then :

$$p^3+2k^6 = a^2$$ and all these variables are integers. You want $p\ne 0$. Take for instance $k = 2$. This rewrites as $$p^3+128 = a^2$$. One solution is $p=17, a=71$.

Therefore a solution to your equation is $({17\over4}+1, {71\over8}) = ({21\over4}, {71\over8})$

To get all the solutions, I suggest reading about Mordell equations to solve $p^3+2k^6=a^2$.

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Elliptic curves can be equipped with a group structure. Let we consider the algebraic curve $y^2=x^3+2$ (that is just a translated version of the original curve). If $A=(x_A,y_A), B=(x_B,y_B)$ are two rational points on such a curve, the line through $A$ and $B$ meet again the curve in a point $(u,v)$ with rational coordinates, as a side-effect of Vieta's theorem. We may define $A\oplus B$ as $C=(u,-v)$ and with such a choice $\oplus$ is associative. We may also define $2A=A\oplus A$ through the intersection between the curve and the tangent through $A$. In particular,

$$ A=(u,v)\quad \implies \quad 2A = \left(\frac{u(u^3-16)}{4v^2},-\frac{u^6+40u^3-32}{8v^3}\right)$$ and by taking $A=(-1,1)$ we get that $2A=\left(\frac{17}{4},-\frac{71}{8}\right)$ is another rational point on the given elliptic curve.

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  • $\begingroup$ Why is this a group ? I recon it is a semigroup, but are there symmetrics and a neutral element ? I'm interrested ... $\endgroup$
    – Astyx
    Nov 4, 2016 at 15:30
  • $\begingroup$ @Astyx: have a look at en.wikipedia.org/wiki/Elliptic_curve $\endgroup$ Nov 4, 2016 at 15:32
  • $\begingroup$ I will. Thanks ! $\endgroup$
    – Astyx
    Nov 4, 2016 at 15:34
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As hinted in the comments, we may consider the equivalent equation $$y^2 = x^3 + 2$$ This is an elliptic curve. Finding rational points is an active domain in research and many books have been written, e.g. Silverman's Rational Points on Elliptic Curves.

Now, you already have some rational points on your curve: $(-1,-1)$ and $(-1,1)$. We can construct new rational points on the elliptic curve by taking the tangent line through one of these points, and calculating the intersection with the elliptic curve.

The tangent line to $(-1,1)$ is $$y = \frac{3}{2}(x+1) + 1 = \frac{3}{2}x+\frac{5}{2}$$ Now, taking this $y$ in the original equation, we get: $$\left(\frac{3}{2}x+\frac{5}{2}\right)^2 = x^3 + 2$$ Solving this cubic for $x$, we obtain two solutions: $x = -1$ and $x = \frac{17}{4}$. You can then calculate $y = \pm \sqrt{\left(\frac{17}{4}\right)^3+2} = \pm \frac{71}{8}$.

So, two more points on your curve are $\left(\frac{17}{4},\frac{71}{8}\right)$ and $\left(\frac{17}{4},-\frac{71}{8}\right)$.

You can then repeat this with the new solution(s) to find more of them, e.g. $$(x,y) = \left(\frac{66113}{80656},\frac{36583777}{22906304}\right)$$

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  • $\begingroup$ Thank you, your solution is much simpler that what I was proposing. $\endgroup$
    – user343447
    Nov 3, 2016 at 22:36

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