$a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$. I was curious if there are quadratic equations where $a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$.
So trivially if $c=0$, $a$ and $b$ can be arbitrary, and if either $a$ or $b$ is zero, this implies that $c=0$, and the other arbitrary.
Is there a way to find other solutions, if any exist?
So the equation to solve in integers is $$a^3b^2c^2+ab^2c+c=0.$$
Dividing out by $c$ gives $$a^3b^2c+ab^2+1=0.$$
We could factor to give $$ab^2(a^2c+1)=-1$$
So either 


*

*$ab^2=1$ and $a^2c+1=-1$  or

*$ab^2=-1$ and $a^2c+1=1$


For 1. $a=1$ and $b=\pm1$ and $c=-2$
For 2. $a=-1$ and $b=\pm1$ and $c=-2$ and $c=0$ is forced.
Is this analysis correct?
Are the non-trivial solutions:
$x^2+x-2$ and $x^2-x-2$, with roots $-2$ and $2$ respectively?
 A: The proof looks correct to me. 
However, it might be good to say "However, we assumed $c \neq 0$" after point 2. 
A: Here's an alternate approach you might consider (+1 on yours, which works just fine).
Now, if $a,b,c$ satisfy the desired property and $c\ne 0,$ then by Rational Root Theorem we have that $abc$ is an integer factor of $c,$ from which we conclude that either


*

*$c=0,$ in which case any $a,b$ will do (as you observed), or

*$c\ne 0$ and $ab$ is an integer factor of $1,$ whence $a,b\in\{1,-1\}$ since $a,b\in\Bbb Z.$


If $c\ne0$ and $a=1,$ then we have by assumption that $$0=(bc)^2+b(bc)+c=b^2c^2+b^2c+c=c^2+2c,$$ whence $c=-2$ since $c\ne 0.$ Hence, $abc=-2b,$ so $$0=(-2b)^2+b(-2b)-2=4b^2-2b^2-2=2b^2-2=2-2,$$ and so either of $b=\pm 1$ works.
If $c\ne 0$ and $a=-1,$ then we have by assumption that $$0=-(bc)^2+b(bc)+c=-b^2c^2+b^2c+c=-c^2+2c,$$ whence $c=2,$ since $c\ne 0.$ Hence, again, $abc=-2b,$ so $$0=-(-2b)^2+b(-2b)+2=-4b^2-2b^2+2=-6b^2+2=-6+2=-4,$$ a contradiction. Hence, we cannot have $a=-1.$
Hence, the solutions are precisely those of the forms $x^2\pm x-2$ or $ax^2+bx$ for some $a,b\in\Bbb Z.$
