
2. The roots of the equation

$$
x^{3}-3 a x^{2}+b x+18 c=0
$$

form a non-constant arithmetic progression and the roots of the equation

$$
x^{3}+b x^{2}+x-c^{3}=0
$$

form a non-constant geometric progression. Given that $a, b, c$ are real numbers, find all positive integral values $a$ and $b$.


## Solution.

Let $\alpha-d, \alpha, \alpha+d(d \neq 0)$ be the roots of the first equation and let $\beta / r, \beta, \beta r(r>0$ and $r \neq 1$ ) be the roots of the second equation. It follows that $\alpha=a, \beta=c$ and

$$
\begin{aligned}
a^{3}-a d^{2} & =-18 c ; \quad 3 a^{2}-d^{2}=b \\
c(1 / r+1+r) & =-b ; \quad c^{2}(1 / r+1+r)=1
\end{aligned}
$$

Eliminating $d, r$ and $c$ yields

$$
a b^{2}-2 a^{3} b-18=0
$$

whence $b=a^{2} \pm(1 / a) \sqrt{a^{6}+18 a}$. For positive integral values of $a$ and $b$ it must be that $a^{6}+18 a$ is a perfect square. Let $x^{2}=a^{6}+18 a$. Then $a^{3}2$ and hence no solution. For $a=1$ there is no solution. For $a=2, x=10$ and $b=9$. Thus the admissible pair is $(a, b)=(2,9)$.
