[sp]If $r=0$ then the equation becomes $2x-1=0$, which does not have a n integer solution. So $r\ne0$ and we can divide through by $r$, getting $x^2 + \bigl(1 + \frac2r\bigr)x + 1 - \frac1r = 0.$ The product of the roots is $1 - \frac1r$, which must be an integer, so $\frac1r$ is an integer, say $r = \frac1n.$ The equation is then $x^2 + (1+2n)x + 1-n = 0$ and its roots are $x = \frac12\bigl(-1-2n \pm\sqrt{(1+2n)^2 - 4(1-n)}\bigr).$ The discriminant is $$(1+2n)^2 - 4(1-n)= 4n^2 + 8n - 3 = 4(n+1)^2 - 7,$$ and this must be a square, say $4(n+1)^2 - 7 = m^2.$ But the only squares that differ by $7$ are $9$ and $16$. It follows that $4(n+1)^2 = 16$, so $n+1 = \pm 2.$ Therefore $n=1$ or $-3$, and $r=1$ or $-\frac13.$ The sum of cubes is therefore $1 - \frac1{27} = \frac{26}{27}.$[/sp]Albert said:r is rational ,and all the roots of equation:
$rx^2+(r+2)x+r-1=0$ are integers
please find :$\sum r^3$
Superb solution, Opalg! (Clapping)Opalg said:[sp]If $r=0$ then the equation becomes $2x-1=0$, which does not have a n integer solution. So $r\ne0$ and we can divide through by $r$, getting $x^2 + \bigl(1 + \frac2r\bigr)x + 1 - \frac1r = 0.$ The product of the roots is $1 - \frac1r$, which must be an integer, so $\frac1r$ is an integer, say $r = \frac1n.$ The equation is then $x^2 + (1+2n)x + 1-n = 0$ and its roots are $x = \frac12\bigl(-1-2n \pm\sqrt{(1+2n)^2 - 4(1-n)}\bigr).$ The discriminant is $$(1+2n)^2 - 4(1-n)= 4n^2 + 8n - 3 = 4(n+1)^2 - 7,$$ and this must be a square, say $4(n+1)^2 - 7 = m^2.$ But the only squares that differ by $7$ are $9$ and $16$. It follows that $4(n+1)^2 = 16$, so $n+1 = \pm 2.$ Therefore $n=1$ or $-3$, and $r=1$ or $-\frac13.$ The sum of cubes is therefore $1 - \frac1{27} = \frac{26}{27}.$[/sp]
Edit. Thanks to Pranav, whose deleted comment set me along the right lines for attacking this problem.