
1. The four real solutions of the equation

$$
a x^{4}+b x^{2}+a=1
$$

form an increasing arithmetic progression. One of the solutions is also a solution of

$$
b x^{2}+a x+a=1 .
$$

Find all possible values of real parameters $a$ and $b$.

(Peter Novotný)


Solution. According to the statement of the problem, the first equation has four distinct solutions, that is $a \neq 0$.

Let $x_{0}$ be the common solution of the equations. Then $x_{0}$ solves the difference of the equations as well, which gives $a x_{0}^{4}-a x_{0}=0$, or $a x_{0}\left(x_{0}^{3}-1\right)=0$. The common solution thus has to be either 0 or 1 .

Substituting $x_{0}=0$ into the first equation gives $a=1$, but the equation $x^{4}+b x^{2}=0$ has the solution 0 with the multiplicity at least two, thus it cannot be $x_{0}=0$.

The common solution therefore is $x_{0}=1$. Substituting this into any of the equations we obtain $b=1-a$ and we can rewrite the first equation as $a x^{4}-(1-2 a) x^{2}+$ $a-1=0$ and we can easily see, that -1 is a solution as well and we have

$$
(x-1)(x+1)\left(a x^{2}-a+1\right)=0 .
$$

The quadratic equation $a x^{2}-(a-1)$ should have two different real solutions, mutually opposite numbers $\xi$ and $-\xi$, which is the case for $a>1$ or $a0$, then if $01$, then the arithmetic progression should be $-\xi,-1,1, \xi$, which gives $\xi=3$. And 3 is a solution of (1), if and only if $a=1 /\left(1-3^{2}\right)=-\frac{1}{8}$, which gives $b=1-2 a=\frac{5}{4}$.

Conclusion. There are two pairs of solutions:

$$
(a, b) \in\left\{\left(-\frac{1}{8}, \frac{5}{4}\right),\left(\frac{9}{8},-\frac{5}{4}\right)\right\}
$$
