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!
Best Answer
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.