Equation with variable roots 
Possible Duplicate:
$f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ 

How many ordered triples of rational numbers $(e,r,t)$ are there such that the following
$$g(x)=x^3+ex^2+rx+t$$
has roots $e$ and $r$ and $t$?
 A: $x^3 + ex^2 +rx + t = (x-e)(x-r)(x-t)$
This gives us the following relations
$$e+r+t = -e \implies r+t+2e = 0$$
$$er+rt+et = r \implies r = \frac{-et}{e+t-1}$$
$$ert = -t \implies er =-1 \quad \mbox{or} \quad t = 0$$
Consider the case $e,r,t\neq 0$:
$e = -\frac{1}{r}$. Putting it in the second equation yields $$t = \frac{r(r+1)}{r^2-1} = \frac{r}{r-1}$$ Substituting values of $e,t$ in first relation yields
$$r+\frac{r}{r-1}-\frac{2}{r} = 0$$
Simplyfying,
$$r^3-2r+2 =0$$
This equation in $r$ has one irrational and two imaginary roots. So, such a solution is not possible.
The other alternatives involve at least one of $e,r,t$ to be $0$.
e = 0
By $ert = -t, t=0. e=0, t=0 \mbox{ gives }r = 0$ by the first relation. So, $(0,0,0)$ is the only solution.
r = 0
$t = 0$ as in the previous case. So, $e=0$ as in the previous case.
t=0
The two non-trivial relations are $er = r, r+2e = 0$. As $r=0$ has already been considered, assume $r\neq 0$. Then, $e = 1, r= -2$. So, (1,-2,0) is a valid solution.
The only two possible solutions are $(0,0,0) \mbox{ and } (1,-2,0)$
