Belarusian Mathematical Olympiad

Answer: $a=2$, $b=4$, $c=-4$.

Let $x_1,x_2$ be distinct roots of the equation $a{x}^2+bx+c=0$, i.e. be the zeroes of the function $f(x)=a{x}^2+bx+c$ and $g(x)=x^3+b{x}^2+ax+c$. By condition, $f(0)=g(0)=c\ne 0$. Let $F(x)=g(x)-f(x)$. Then $0,x_1,x_2$ are the distinct zeroes of the polynomial $F(x)$. So, $F(x)=x(x-x_1)(x-x_2)$, then $$\begin{gathered} x^3+(b-a)x^2+(a-b)x=x(x-x_1)(x-x_2)\Rightarrow \\ (b-a)x+(a-b)=-(x_1+x_2)x+x_1{x}_2\forall x\in \mathbb{R}. \end{gathered}$$ Therefore, $x_1+x_2=a-b$ and $x_1{x}_2=a-b$. By the Vieta theorem $x_1+x_2=-b/a$, $x_1{x}_2=c/a$, hence $c=-b$. Further, $a-b=-b/a$ or $a^2-ab=-b$, i.e. $a^2=b(a-1)$. Note that we have $a\ne 1$, so $$b=\frac{a^2}{a-1}=\frac{(a^2-1)+1}{a-1}=a+1+\frac{1}{a-1}.$$ Since $a$ and $b$ are integers, we see that $1/(a-1)$ is also integer, which is possible only if $a=0$ and $a=2$. But $a\ne 0$, so $a=2$ and then $b=4$, $c=-4$.

Straightforward calculation show that the obtained triple satisfy the problem condition. Indeed, the equation $2x^2+4x-4=0$ has two distinct roots $-1\pm \sqrt{3}$. These numbers are the roots of the equation $x^3+4x^2-2x-4$ since $$\begin{gathered} (-1\pm \sqrt{3}{)}^3+4\cdot (-1\pm \sqrt{3}{)}^2-2\cdot (-1\pm \sqrt{3})-4= \\ =-1\pm 3\sqrt{3}-3\cdot 3\pm 3\sqrt{3}+4\mp 8\sqrt{3}+12-2\pm 2\sqrt{3}-4=0. \end{gathered}$$