Find all ordered triplets $(p,q,r)$

Question

Let $A,G^2,H$ be the roots of the cubic equation $x^3+px^2+qx+r=0$ which are in G.P., where $p,q$ are integers and $A,G,H$ are respectively AM,GM,HM of two positive numbers.
If $p,q \in (-100,100)$, then find the number of all possible ordered triplets $(p,q,r)$.
I tried writing sum of roots...., but it made a lot of cases ($r$ is negative, etc). Some hints please. Thanks.

Answer 1

Let $\color{Green}{x_1=\dfrac12(a+b)},\,\,\,\color{Green}{x_2=ab}$ and $\color{Green}{x_3=\dfrac{2ab}{(a+b)}}$ be the roots of given cubic equation.
If they are in a geometric progression, we have $$x_2^2=x_1x_3$$ $$a^2b^2=\dfrac12(a+b)\times\dfrac{2ab}{(a+b)}=ab$$ $$ab=0\,\,\text{or}\,\,\ ab=1.$$

CASE 1: $ab=0$
$x_2=x_3=0$
$q=r=0$ and $p$ can be any number in the given range.
($201$ possible triples, including both $-100$ and $100$.)

CASE 2: $ab=1$
$x_2=1$
$r=-x_1x_2x_3=-1$ and this implies $1+p+q-1=0.$
Hence $p, q$ can be any numbers in the given range satisfying $p+q=0$.
(Again we have $201$ possible triples.)

Hence there are $\color{Red}{402}$ such $(p, q, r)$ distinct ordered triples.