Reduce me non-constantly

We first consider that $f_n (x)$ has linear factor(s): by Rational Root Theorem, possible rational roots of $f_n (x)$ is $\pm 1, \pm 3, \pm 9$ . So at least one $f_n (\pm 1) , f_n (\pm 3) , f_n (\pm 9)$ equals to $0$ for some integer $n$ . It's trivial that $f_n (-1) = 0$ and $f_n (1) = 0$ satisfy integer $n$ , which gives $n=11, -13$ . If $f_n (\pm 3) = 0 \Rightarrow \pm \space 3^5 \pm \space 3^3 n - 3^3 - 9 =0 \Rightarrow n$ is not an integer. Similarly if $f_n (\pm 9) = 0 \Rightarrow \pm \space 9^5 \pm \space 9^3 n - 3 \cdot 9^2 - 9 =0 \Rightarrow n$ is not an integer.

So if there's any more possible values of $n$ , $f_n (x)$ can be be factored in terms of two irreductible polynomials: $(x^2 + ax + b) (x^3 + cx^2 + dx + e)$ , where the $a,b,c,d,e$ are integers.

$(x^2 + ax + b) (x^3 + cx^2 + dx + e) = x^5 + nx^3 - 3x^2 - 9$

Expand and compare distinct powers of $x$ :

$a + c = 0 \Rightarrow a = -c$

$b + d - a^2 = n$ , $(*)$

$a = \frac {e+3}{b-d}$

$a = - \frac {bd}{e}$

Or $a = - \frac {bd}{e} = \frac {e+3}{b-d}$ , $(**)$

$be = -9$ , $(***)$

From $(***)$ , we have all possible cases: $(b,e) = (1,-9), (3,-3),(9,-1), (-1,9),(-3,3),(-9,1)$

If $(b,e) = (1,-9)$ , from $(**)$ , there's no integer solution.

If $(b,e) = (3,-3)$ , from $(**)$ , $a=d=0 \Rightarrow n = b+d-a^2 = 3$

If $(b,e) = (9,-1)$ , from $(**)$ , there's no integer solution.

If $(b,e) = (-1,9)$ , from $(**)$ , there's no integer solution.

If $(b,e) = (-3,3)$ , from $(**)$ , $a=d=3 \Rightarrow n = b+d-a^2 = -3$

If $(b,e) = (-9,1)$ , from $(**)$ , there's no integer solution.

Hence the answer is $|11| \times |-13| \times |3| \times |-3| \bmod{1000} = \boxed{287}$

Let $f = g \times h$ . Degree of f is 5.

Firstly, we are looking for polynomial g with degree 1, that means g = x-a, with f(a) = 0.

Because of integer coefficients, the value of a shall be divisor of 9 (the free coefficient).

For a = 1, f(1) = 0 and we obtain n = 11

For a= -1, f(-1) = 0 and n= -13

For values 3, -3, 9, -9 of a , we don't have integers n.

Now, we are looking for g and h with degrees 2 and 3:

$f = X^{5} + n X^{3} - 3 X^{2} - 9$

$f = (X^{2} + a X + b) \times (X^{3} + c X^{2} + d X + e)$ .

After making this product we obtain:

a+c=0

$b+d - a^{2} =n$

a d - a b + e = - 3

a e + b d = 0

and b e = - 9.

We'll discuss all the cases for the last condition b e = - 9 with b and e integer numbers !

For the (b, e) pairs (-9, 1), (-3, 3), (-1, 9) and (9, -1) we don't have integer solutions for a or d.

But for the pair b=3 and e = -3 we found

a = d = 0, so n = 3 (and c = 0)

or a = d = 3, so n= - 3 (and c = - 3)

The product of absolute values of n is 11 * 13* 3 * 3 = 1287, so the last three digits are 287

We can now admire all the four possible polynomial factorizations:

$X^{5} + 11 X^{3} - 3 X^{2} - 9 = (X - 1) \times (X^{4} + X^{3} + 12 X^{2} + 9 X + 9)$

$X^{5} - 13 X^{3} - 3 X^{2} - 9 = (X + 1) \times (X^{4} - X^{3} - 12 X^{2} + 9 X - 9)$

$X^{5} + 3 X^{3} - 3 X^{2} - 9 = (X^{2} + 3) \times (X^{3} - 3)$ .

$X^{5} - 3 X^{3} - 3 X^{2} - 9 = (X^{2} + 3 X + 3) \times (X^{3} - 3 X^{2} + 3 X - 3)$

Good luck for the next week of Brilliant challenges :)

$f_n(x)$ can be rewritten as $(x^4+ax^3+bx^2+cx+d)(x+e)$ . Since this has to be equal with the original $f_n(x)$ one obtains five equations: $e+a=0$ , $ea+b=n$ , $be+c=-3$ , $ce+d=0$ and $de=-9$ .

Since the coefficients have to be integers, there are only six possible values for d and e . The equation $ce+d=0$ has to be met, therefore $d=-1$ and $e=9$ and also $d=1$ and $e=-9$ do not fit.

As well the equation $be+c=-3$ has to be met, that means that $(c, d, e) \neq(1, -3, 3)$ and $\neq (1, 3, -3)$ .

With $e=-a$ there are only two possible autcomes for $n=ae+b$ are $-13$ and 11.

But $f_n(x)$ can also be rewritten as $(x^3+ax^2+bx+c)(x^2+dx+e)$ . Solving the equation as above, there are only two outcomes $(a, b, c, d, e, n)=(0, 0, -3, 0, 3, 3)$ or $(-3, 3, -3, 3, 3, -3)$ .

The muliplication of the absolute value of the four possibilities for n is $3\times-3\times11\times-13=1287$ . Therefore the answer is 287.