MHB Finding the Sum of Cubes for Rational Numbers with Integer Roots

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The discussion focuses on finding the sum of cubes of rational numbers \( r \) such that all roots of the equation \( rx^2 + (r+2)x + r-1 = 0 \) are integers. It is established that \( r \) cannot be zero, leading to the transformation of the equation into a form where the product of the roots must be an integer. By setting \( r = \frac{1}{n} \) and analyzing the discriminant, it is determined that the only valid values for \( n \) are 1 and -3, resulting in \( r = 1 \) or \( r = -\frac{1}{3} \). Consequently, the sum of cubes is calculated as \( \frac{26}{27} \). The discussion concludes with acknowledgment of contributions that guided the solution process.
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r is rational ,and all the roots of equation:
$rx^2+(r+2)x+r-1=0$ are integers
please find :$\sum r^3$
 
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Albert said:
r is rational ,and all the roots of equation:
$rx^2+(r+2)x+r-1=0$ are integers
please find :$\sum r^3$
[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.
 
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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.
Superb solution, Opalg! (Clapping)
 
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