Croatian Mathematical Competitions

First, expand both sides:

Left side:
$(a - 1)(1 + x + x^2)^2 = (a - 1)(1 + 2x + 3x^2 + 2x^3 + x^4)$

Right side:
$(a + 1)(1 + x^2 + x^4)$

Bring all terms to one side:
$$(a - 1)(1 + 2x + 3x^2 + 2x^3 + x^4) - (a + 1)(1 + x^2 + x^4) = 0$$

Expand:
$(a - 1)(1) + (a - 1)(2x) + (a - 1)(3x^2) + (a - 1)(2x^3) + (a - 1)(x^4)$
$- (a + 1)(1) - (a + 1)(x^2) - (a + 1)(x^4) = 0$

Group like terms:
- Constant: $(a - 1) - (a + 1) = -2$
- $x$ term: $2(a - 1)$
- $x^2$ term: $3(a - 1) - (a + 1) = 3a - 3 - a - 1 = 2a - 4$
- $x^3$ term: $2(a - 1)$
- $x^4$ term: $(a - 1) - (a + 1) = -2$

So the equation becomes:
$$-2 + 2(a - 1)x + (2a - 4)x^2 + 2(a - 1)x^3 - 2x^4 = 0$$

Divide both sides by $2$:
$$-1 + (a - 1)x + (a - 2)x^2 + (a - 1)x^3 - x^4 = 0$$

Or:
$$-x^4 + (a - 1)x^3 + (a - 2)x^2 + (a - 1)x - 1 = 0$$

Or, multiplying both sides by $-1$:
$$x^4 - (a - 1)x^3 - (a - 2)x^2 - (a - 1)x + 1 = 0$$

Let us factor this quartic. Notice that if $x = 1$:

$x^4 - (a - 1)x^3 - (a - 2)x^2 - (a - 1)x + 1 = 1 - (a - 1) - (a - 2) - (a - 1) + 1$
$= 1 - a + 1 - a + 2 - a + 1 + 1$
But let's compute step by step:
$1 - (a - 1) = 2 - a$
$2 - a - (a - 2) = 2 - a - a + 2 = 4 - 2a$
$4 - 2a - (a - 1) = 4 - 2a - a + 1 = 5 - 3a$
$5 - 3a + 1 = 6 - 3a$
So $x = 1$ is a root if $6 - 3a = 0$, i.e., $a = 2$.

Try $x = -1$:
$(-1)^4 - (a - 1)(-1)^3 - (a - 2)(-1)^2 - (a - 1)(-1) + 1$
$= 1 - (a - 1)(-1) - (a - 2)(1) - (a - 1)(-1) + 1$
$= 1 + (a - 1) - (a - 2) + (a - 1) + 1$
$= 1 + a - 1 - a + 2 + a - 1 + 1$
$= (1 - 1 + 2 + 1) + (a - a + a) - 1$
$= (3) + (a) - 1$
$= a + 2$
So $x = -1$ is a root if $a = -2$.

Alternatively, factor the quartic as follows:

Let us try to factor as $(x^2 + bx + 1)(x^2 + cx + 1)$.

Expand:
$(x^2 + bx + 1)(x^2 + cx + 1) = x^4 + (b + c)x^3 + (bc + 2)x^2 + (b + c)x + 1$

Compare with $x^4 - (a - 1)x^3 - (a - 2)x^2 - (a - 1)x + 1$:

So:
- $x^4$ coefficient: $1$
- $x^3$ coefficient: $b + c = -(a - 1)$
- $x^2$ coefficient: $bc + 2 = -(a - 2)$
- $x$ coefficient: $b + c = -(a - 1)$
- constant: $1$

So $b + c = -(a - 1)$
$bc + 2 = -(a - 2)$

Let $b + c = s = -(a - 1)$
$bc = -(a - 2) - 2 = -(a) + 2 - 2 = -a$

So $b$ and $c$ are roots of $t^2 - s t - a = 0$

So $t^2 + (a - 1)t - a = 0$

Thus, the quartic factors as:
$$(x^2 + b x + 1)(x^2 + c x + 1) = 0$$
where $b$ and $c$ are roots of $t^2 + (a - 1)t - a = 0$.

Therefore, the solutions are all real $x$ such that
$$x^2 + b x + 1 = 0 \quad \text{or} \quad x^2 + c x + 1 = 0$$
where $b$ and $c$ are roots of $t^2 + (a - 1)t - a = 0$.

Explicitly, $b, c = \frac{-(a - 1) \pm \sqrt{(a - 1)^2 + 4a}}{2} = \frac{-(a - 1) \pm \sqrt{a^2 + 2a + 1 + 4a}}{2} = \frac{-(a - 1) \pm \sqrt{a^2 + 6a + 1}}{2}$

So the solutions are all real $x$ such that
$$x^2 + \left(\frac{-(a - 1) + \sqrt{a^2 + 6a + 1}}{2}\right)x + 1 = 0$$
or
$$x^2 + \left(\frac{-(a - 1) - \sqrt{a^2 + 6a + 1}}{2}\right)x + 1 = 0$$

That is, for each $a$, the equation has at most four real solutions, given by the roots of these two quadratics.

Special cases:
- If $a = 2$, then $b + c = -(2 - 1) = -1$, $bc = -2$; so $b$ and $c$ are roots of $t^2 + t - 2 = 0$, i.e., $t = 1, -2$.
So the quadratics are $x^2 + x + 1 = 0$ and $x^2 - 2x + 1 = 0$.
The first has no real roots, the second is $(x - 1)^2 = 0$, so $x = 1$ is a double root.
- If $a = -2$, then $b + c = -(-2 - 1) = 3$, $bc = 2$; so $t^2 - 3t - 2 = 0$, $t = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$.
Both quadratics $x^2 + b x + 1 = 0$ and $x^2 + c x + 1 = 0$ have real roots.

In summary:

For each real $a$, the solutions are all real $x$ such that
$$x^2 + b x + 1 = 0 \quad \text{or} \quad x^2 + c x + 1 = 0$$
where $b$ and $c$ are the roots of $t^2 + (a - 1)t - a = 0$.