Problem 9.2. Let $b$ and $c$ be real numbers such that the equation $x^{2}+b x+c=$ 0 has two distinct real roots $x_{1}$ and $x_{2}$ with $x_{1}=x_{2}^{2}+x_{2}$.

a) Find $b$ and $c$ if $b+c=4$.

б) Find $b$ and $c$ if they are coprime integers.

Stoyan Atanasov
9.2. It follows from the condition $x_{1}=x_{2}^{2}+x_{2}$ and Vieta's formulas that

$$
\left\lvert\, \begin{aligned}
x_{1}+(b-1) x_{2} & =-c \\
x_{1}+x_{2} & =-b \\
x_{1} x_{2} & =c
\end{aligned}\right.
$$

Hence $c^{2}+4(1-b) c+b^{3}-b^{2}=0, b \neq 2$.

a) Since $c=4-b$, we obtain $b^{3}+4 b^{2}-28 b+32=0$ which is equivalent to $(b-2)^{2}(b+8)=0$. Therefore $b=-8$ and $(b, c)=(-8,12)$.

b) Consider $c^{2}+4(1-b) c+b^{3}-b^{2}=0$ as a quadratic equation of $c$. It follows that $D=16(1-b)^{2}-4\left(b^{3}-b^{2}\right)=4(1-b)(b-2)^{2}$ is a perfect square. Thus $b=2$ or $1-b=k^{2}$, where $k$ is an integer. Then $(b, c)=(2,2)$ or $(b, c)=\left(1-k^{2}, k(k-1)^{2}\right)$. Obviously the first pair is not a solution of the problem. The integers in the second pair are coprime when $k-1= \pm 1$, that is, $k=2$ or $k=0$. So $(b, c)=(-3,2)$ or $(b, c)=(1,0)$. In both cases the roots of the given equation are real and distinct.