# Cubic diophantine equation

How can be solved the equation $x^3+x-1=y^2$ in positive integers?

I know this equation defines an elliptic curve, but this seems to be a non-elementary way to solve this question. Is there a more elementary solution?

By the way I found three solutions: $(1,1), (2,3)$ and $(13,47)$. Is this related to the law group property of elliptic curves? Thanks for further answers.

• The only thing I can think of is $(x-1)(x^2 + x + 2) = (y + 1)(y - 1)$ – Yiyuan Lee Mar 7 '14 at 16:01
• $x(x^2+1)=y^2+1$. Now notice that x and $1+x^2$ are coprime. – Lucian Mar 7 '14 at 16:03
• Thanks, but in this form how can the equation be solved? Maybe using algebraic number theory. The problem is that $y^2+1$ is not a square, cubic, etc. – Xam Mar 7 '14 at 16:07
• To find integral solutions for $x^3+x-1=y^2$ see also math.stackexchange.com/questions/32847/…. – Dietrich Burde Mar 7 '14 at 16:17
• @DietrichBurde Thanks so much. For this problem I just recently introducing me to this topic of elliptic curves. – Xam Mar 7 '14 at 16:25

Perhaps others can find an elementary solution. This is an elliptic curve, of rank $1$ and trivial torsion subgroup. So the group of all rational points is generated by one single rational point, namely $P=(1,1)$. Hence, every rational point on the curve is of the form $nP$ for some $n\in\mathbb{Z}$. Here are some multiples: $$2P=(2 , -3 ),\ 3P=(13 , 47),\ 4P=\left(\frac{25}{36} , \frac{37}{216}\right),\ 5P=\left(\frac{685}{121} , -\frac{18157}{1331}\right),\ldots$$ Some cumbersome arguments using heights can show that, in fact, the point at infinity, together with $\pm P$, $\pm 2P$, and $\pm 3P$ are the only integral points on the curve (since this curve is of the form $y^2=f(x)$, if $Q=(x_0,y_0)$ is a point on the curve, then $-Q=(x_0,-y_0)$).