Problem 1663

1. Given the polynomial equation:
$$(x^3 + 4x^2 + 4x + 3)f(x) = (x^3 - 2x^2 + 2x - 1)f(x+1)$$
we need to find the polynomial $f(x)$ with integer coefficients and prove that for each positive integer $n \geq 8$, $f(n)$ has at least five distinct prime divisors.

2. First, factorize the polynomials on both sides of the equation:
$$x^3 - 2x^2 + 2x - 1 = (x-1)(x^2 - x + 1)$$
$$x^3 + 4x^2 + 4x + 3 = (x+3)(x^2 + x + 1)$$
Thus, the equation becomes:
$$(x+3)(x^2 + x + 1)f(x) = (x-1)(x^2 - x + 1)f(x+1)$$

3. Substitute $x = 1$ into the equation:
$$(1+3)(1^2 + 1 + 1)f(1) = (1-1)(1^2 - 1 + 1)f(2)$$
$$4 \cdot 3 f(1) = 0 \implies f(1) = 0$$

4. Substitute $x = -3$ into the equation:
$$(-3+3)((-3)^2 - 3 + 1)f(-3) = (-3-1)((-3)^2 + 3 + 1)f(-2)$$
$$0 \cdot 7 f(-3) = -4 \cdot 13 f(-2) \implies f(-2) = 0$$
Therefore, $f(x)$ has roots at $x = 1$ and $x = -2$. Thus, $f(x) = (x-1)(x+2)g(x)$ for some polynomial $g(x)$.

5. Substitute $f(x) = (x-1)(x+2)g(x)$ into the original equation:
$$(x+3)(x^2 + x + 1)(x-1)(x+2)g(x) = (x-1)(x^2 - x + 1)(x+1)(x+2)g(x+1)$$
Canceling common factors:
$$(x+3)(x^2 + x + 1)g(x) = (x^2 - x + 1)xg(x+1)$$

6. Substitute $x = 0$ into the simplified equation:
$$(0+3)(0^2 + 0 + 1)g(0) = (0^2 - 0 + 1)0g(1)$$
$$3 \cdot 1 g(0) = 0 \implies g(0) = 0$$

7. Substitute $x = -2$ into the simplified equation:
$$(-2+3)((-2)^2 + (-2) + 1)g(-2) = ((-2)^2 - (-2) + 1)(-2)g(-1)$$
$$1 \cdot 1 g(-2) = 0 \implies g(-1) = 0$$
Therefore, $g(x) = x(x+1)h(x)$ for some polynomial $h(x)$.

8. Substitute $g(x) = x(x+1)h(x)$ into the simplified equation:
$$(x^2 + x + 1)h(x) = (x^2 - x + 1)h(x+1)$$

9. Since the roots of $x^2 - x + 1$ and $x^2 + x + 1$ are disjoint, $h(x)$ must be of the form $h(x) = (x^2 - x + 1)q(x)$ for some polynomial $q(x)$.

10. Substitute $h(x) = (x^2 - x + 1)q(x)$ into the equation:
$$q(x) = q(x+1)$$
This implies $q(x)$ is a constant polynomial. Let $q(x) = c$, where $c \in \mathbb{Z} \backslash \{0\}$.

11. Therefore, the polynomial $f(x)$ is:
$$f(x) = c(x-1)x(x+1)(x+2)(x^2 - x + 1)$$

12. For $n \geq 8$, consider $f(n)$:
$$f(n) = c(n-1)n(n+1)(n+2)(n^2 - n + 1)$$
Since $n, n-1, n+1, n+2$ are four consecutive integers, they are pairwise coprime, and each has at least one distinct prime divisor. Additionally, $n^2 - n + 1$ is not divisible by any of $n, n-1, n+1, n+2$ and thus provides at least one more distinct prime divisor.

13. Therefore, $f(n)$ has at least five distinct prime divisors for $n \geq 8$.

$\blacksquare$