Help on solutions of the congruence $f(x)=x^3+4x+8 \equiv 0 \pmod {15}$ I'm doing a little exercise, solve the congruence $f(x)=x^3+4x+8 \equiv 0 \pmod {15}$.
I know that $15=3 \times 5$ and they are relatively prime, so I can split the congruence into:
a) $f(x) \equiv 0 \pmod {3}$
b) $f(x) \equiv 0 \pmod {5}$
I proceed now with the individual solutions.
a) I can reduce the function because I know $x^3 \equiv x \pmod 3$, so it became $5x \equiv -8 \pmod 3 \Rightarrow x \equiv 2 \pmod 3$ and this is one solution
b) I can't reduce because the degree of the monic polynomial $f(x)$ is less than $x^5 \equiv x \pmod 5$. If I try all the values of $x \in [0,1,2,3,4]$ into $f(x)$, I can't obtain any zero $(x,f(x))\rightarrow (0,8),(1,13),(2,24),(3,47),(4,88)$.
So seems like there is no solution.
At this point my question is: the only solution is that from the point a)? Or maybe I'm missing something or maybe I'm wrong!
 A: You're right. There are no solutions.
Note that modulo $5$ we have
$$
x^3 + 4x + 8 \equiv x^3-x + 3 = x(x-1)(x+1) + 3
$$
meaning we can exclude $x = 0, 1, 4$ immediately, and making the calculation for $n = 2, 3$ a bit easier.
As an additional check, you can enter it into WolframAlpha (press "more" next to the table to get a table with all $15$ possible inputs) and see that indeed, there are no solutions.
A: An additional check is also that the polynomial $x^3+4x+3$ is irreducible over $\Bbb F_5$ by the Berlekamp algorithm. So there is no root. Over $\Bbb F_3$ we obtain
$$
x^3+4x+8=(x^2 + 2x + 2)(x + 1),
$$
where $x^2+2x+2$ is irreducible. So the only root is $x=2$.
A: As other answrers note, there are no solutions because the equation fails $\bmod 5$.
Here an alternate method for solving cubic equations $\bmod 5$ is explored.  Suppose the equation has the form
$ax^2+bx^2+cx+d\equiv 0\bmod 5$  Eq. 1
If $d\equiv 0$ then $x\equiv 0$ is a root and the quadratic equation $ax^2+bx+c=0$ may be solved for the other root bystanders techniques for quadratic equations.
Otherwise multiply by $ax-b$ to obtain a quartic equation
$a^2x^4+(ac-b^2)x^2+(ad-bc)x-bd\equiv 0$
and then, since $d\not\equiv 0$ forces $x\not\equiv 0$, we must have $x^4\equiv 1$.  Thereby
$(ac-b^2)x^2+(ad-bc)x+(a^2-bd)\equiv 0$  Eq. 2
Or perhaps easier to remember:
$a(cx^2+dx+a)-b(bx^2+cx+d)\equiv 0$  Eq. 2a
Note that any zero root to this equation must be rejected as inconsistent with the cubic equation having $d\not\equiv 0$, and a root satisfying $ax=b$ must be checked against the cubic equation because multiplying the original cubic by $ax-b$.  Apart from these checks Eq. 2 or 2a will be equivalent to the original cubic.
For the problem at hand we have $x^3+4x+8\equiv 0$.  Then from Eq. 2a we see immediately that $4x^2+8x+1=0$.  Multiplying by $4$ to render the equation monic and completing the square gives $(x+1)^2\equiv 2$, which has no solution as $2$ is not a quadratic residue $\bmod 5$.  So the cubic equation also fails.
