## 1. Roots of the Equation

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

$v$ in the set of real numbers are four consecutive terms of an increasing arithmetic sequence. One of these terms is also a solution to the equation

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

Determine all possible values of the real parameters $a, b$.
SOLUTION. Given that the increasing arithmetic sequence is formed by four distinct real numbers, the first of the given equations must have four distinct real roots. Therefore, $a \neq 0$.

Let $x_{0}$ be the common root of both equations. Then $x_{0}$ is also a root of the equation obtained by subtracting the second given equation from the first, i.e., the equation $a x^{4} - a x = 0$. This can be further simplified to $a x (x^{3} - 1) = 0$. For the common real root $x_{0}$ of both given equations, it follows that $x_{0} = 0$ or $x_{0} = 1$.

Substituting $x_{0} = 0$ into the first given equation, we get $a = 1$, so the equation is of the form $x^{4} + b x^{2} = 0$. However, this equation does not have four distinct real roots for any real number $b$ (0 is at least a double root), hence $x_{0} \neq 0$.

The only common root of both equations is therefore $x_{0} = 1$. Substituting this value into either of the given equations, we get $b = 1 - 2a$. The first equation can then be written as $a x^{4} + (1 - 2a) x^{2} + a - 1 = 0$, from which it is clear that it also has the root -1, and after factoring out the product of the root factors $(x-1)(x+1)$, we get the equation

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

The quadratic polynomial $a x^{2} - (a - 1)$ must have two distinct roots, which must be two mutually opposite (non-zero) numbers $\xi$ and $-\xi$. This is satisfied precisely when $(a-1) / a > 0$, i.e., precisely when $a > 1$ or $a < 0$. For $a < 0$, we get two possibilities for the arithmetic sequence of all four roots depending on whether $0 < a < 1$ or $a > 1$.

In the first case, the four roots of the equation (1) form the arithmetic sequence $-1, -\xi, \xi, 1$, which clearly has a difference of $\frac{2}{3}$, so $\xi = 1 - \frac{2}{3} = \frac{1}{3}$. This number $\xi$ is a root of equation (1) if and only if $a = 1 / (1 - \xi^{2}) = \frac{9}{8}$. Then $b = 1 - 2a = -\frac{5}{4}$.

In the second case, the four roots of the equation (1) form the arithmetic sequence $-\xi, -1, 1, \xi$ with a difference of 2, so $\xi = 1 + 2 = 3$. The number 3 is a root of equation (1) if and only if $a = 1 / (1 - 3^{2}) = -\frac{1}{8}$. Then $b = 1 - 2a = \frac{5}{4}$.

Conclusion. The problem is satisfied by exactly two pairs of real numbers $(a, b)$, namely

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

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Determine all values of the real parameters $p$ and $q$ for which each of the equations

$$
x(x-p) = 3 + q, \quad x(x+p) = 3 - q
$$

has two distinct roots in the set of real numbers, the arithmetic mean of which is one of the roots of the other equation. [59-B-S-1]

N2. Determine all pairs $(a, b)$ of real numbers for which each of the quadratic equations

$$
a x^{2} + 2 b x + 1 = 0, \quad b x^{2} + 2 a x + 1 = 0
$$

has two distinct real roots, one of which is common to both equations.

N3. Determine all pairs $(a, b)$ of real numbers for which the equations

$$
x^{2} + (3a + b)x + 4a = 0, \quad x^{2} + (3b + a)x + 4b = 0
$$

have a common real root. [57-B-S-2]

D1. Determine all three-term arithmetic sequences of prime numbers with a difference of 1970.

$[20-\mathrm{B}-\mathrm{P}-2]$

D2. Consider the two quadratic equations

$$
x^{2} - a x - b = 0, \quad x^{2} - b x - a = 0
$$

with real parameters $a, b$. Determine the smallest and the largest values that the sum $a + b$ can take if there exists exactly one real number $x$ that satisfies both equations. Also determine all pairs $(a, b)$ of real parameters for which the sum of these values is achieved. [57-B-II-1]

D3. The real numbers $a, b$ have the following property: the quadratic equation $x^{2} - a x + b - 1 = 0$ has two distinct roots in the set of real numbers, the difference of which is a positive root of the equation $x^{2} - a x + b + 1 = 0$.

a) Prove the inequality $b > 3$.

b) Express the roots of both equations in terms of $b$.

$[59-\mathrm{B}-\mathrm{I}-6]$