$y^{2}=(x-1)^{2}(x+2)$ $y=(x-1)\sqrt{x+2}$ So we need x+2 to be a perfect square. (2,2) (7,18 ) (this was a smiley earlier, hence the space) (14,52) (23,110) There are infinitely many.

Moonmathpi496 wrote: You found integer solution.But what about rational solutions? Example: 1/4,7/9 are some solutions. Thangtonghop wrote: I think to change to Rational roots, we find p/q+2= perfect rational square. It 's easy because (p,q)=1 so p+2q, q=perfect squares. We solve x^2- 2y^2= p , which is Pell equation

bpms wrote: $ y^{2} = (x - 1)^{2}(x + 2)$ $ y = (x - 1)\sqrt {x + 2}$ So we need x+2 to be a perfect square. (2,2) (7,18 ) (this was a smiley earlier, hence the space) (14,52) (23,110) There are infinitely many. I think solution : $ x = m^2 - 2,m\in Q$ $ |y| = |(m^2 - 3)m|$

TTsphn wrote: bpms wrote: $ y^{2} = (x - 1)^{2}(x + 2)$ $ y = (x - 1)\sqrt {x + 2}$ So we need x+2 to be a perfect square. (2,2) (7,18 ) (this was a smiley earlier, hence the space) (14,52) (23,110) There are infinitely many. I think solution : $ x = m^2 - 2,m\in Q$ $ |y| = |(m^2 - 3)m|$ I think , when x=1 , x+2 need not a perfect square. and x=1 => y=0 is solution. but x=1=m^2-2 , m=3^(1/2) is not rational number. so, m is also include 3^(1/2) my english is very poor. I do not well english. sorry. it is first time I write reply in ML

\[ y=(x-1)\sqrt{x+2}\] is rational. Thus, either $ \sqrt{x+2}$ is rational or $ x-1$ is zero. Thus the solution set is: \[ (x,y)=(1,0)\qquad (x,y)=\left(m^2-2,m(m^2-3)\right),\qquad m\in\mathbb{Q}\]