<s>[INST] find all possible triples of real numbers $(a, b, c)$. 
Given the real numbers $a, b, c$ so that the equation
$$x^2 + ax + b = 0$$
has exactly two different solutions, namely $p$ and $q$, and the
equation
$$x^3 + ax^2 + bx +c= 0$$
has exactly two distinct solutions, namely $p_0$ and $q_0$. If it
turns out that $p_0- p = q_0 -q$, find all possible triples of real
numbers $(a, b, c)$.

Suppose
$f(x)=x^3+ax^2+bx+c$ and $g(x)=x^2+ax+b$
Obviusly, $f(p)=f(q)=c$
$WLOG$, $p_0$ is double root of $f(x)$
Using Vieta
$$p+q=-a$$
$$pq=b$$
$$2p_0+q_0=-a$$
$$p_0^2+2p_0q_0=b$$
$$p_0^2q_0=-c$$
Observe that another form of $f(x)$ and $g(x)$
$f(x)=(x-p_0)^2(x-q_0)$ and $g(x)=(x-p)(x-q)$
$$f(p)-f(q)=(p-p_0)^2(p-q_0)-(q-p_0)^2(q-q_0)$$
$$0=(q-q_0)[(p-p_0)(p-q_0)-(q-p_0)^2]$$
$q-q_0=0$ or $(p-p_0)(p-q_0)-(q-p_0)^2=0$

For $q-q_0=0$

So we get that $p=p_0$, $q=q_0$ and $c=0$. Cause $c=0$, $f(x)=x(x^2+ax+b)$. So the roots of $f(x)$ is $p, q, 0$, since $p-q\neq 0$ so $p=0$ and $q=-a$, caused $b=0$. $(a, b, c)=(a, 0, 0)$ for $q\neq 0$

I'm confused how to use $(p-p_0)(p-q_0)-(q-p_0)^2=0$

 [/INST] WLOG, $p_0$ is double root of $x^3+ax^2+bx+c$, then $2p_0+q_0=-a=p+q \Rightarrow p_0+(p_0-p)+(q_0-q)=0$. $p_0-p=q_0-q \Rightarrow p_0+2(p_0-p)=0 \Rightarrow p=\frac32 p_0 \Rightarrow q=q_0+\frac{p_0}2$.
Then $b=pq=\frac32 p_0q_0+\frac34 p_0^2$. Also $b=2p_0q_0+p_0^2$. Then $\frac32 p_0q_0+\frac34 p_0^2=2p_0q_0+p_0^2 \Rightarrow$ $\frac12 p_0q_0+ \frac14 p_0^2 \Rightarrow$ $2p_0q_0+p_0^2=0 \Rightarrow b=0$.
Then one of $p$ and $q$ is 0.

*

*$p=0 \Rightarrow p_0=0\Rightarrow c=0$. Then $q=q_0=-a$ and $a$ is free parameter. $(a,b,c)=(a,0,0)$ with $a\neq 0$.


*$q=0 \Rightarrow p=-a \Rightarrow p_0=-\frac23a \Rightarrow q_0=q-\frac{p_0}2=\frac13a \Rightarrow c=-p_0^2q_0=-\frac4{27}a^3$. $(a,b,c)=(3t,0,-4t^3)$ with $t=\frac{a}3\neq 0$.
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