Question: Find all the solutions of the polynomial congruence $$f(x)=x^4+x^3+x+10 \equiv 0\pmod {40}$$
Answer: We have that $40=2^3\cdot 5$. So, the polynomial congruence is equivalent with the system $$x^4+x^3+x+10 \equiv 0\pmod 5,\ x^4+x^3+x+10 \equiv 0\pmod{2^3}$$
The only solution for the first is $x\equiv 0 \pmod 5$. For the second we have that $x^4+x^3+x+10 \equiv 0\pmod{2^3}$ is in the form of $f(x) \equiv 0 \pmod{p^3}$, where $p$ is a prime, $f(x)=x^4+x^3+x+2$, $f'(x)=4x^3+3x^2+1.$
- The class $0\pmod 2$ is a solution for $f(x) \equiv 0 \pmod{2}$, and $2\nmid f'(0)=1. $ So, one solution for the congruence $f(x)\equiv 0 \pmod{2^2}$ is the class $0+t\cdot2\pmod{2}$, where $t\in \mathbb{Z}: f(0)/2+t\cdot f'(0) \equiv 0 \pmod{2}\iff t\equiv 1\pmod{2}$. To sum up, we found that the class $ 2 \pmod{2^2}$ is a solution for $f(x) \equiv 0 \pmod{2^2}$.
- Similarly, $2\pmod{2^2}$ is a solution for $f(x) \equiv 0 \pmod{2^2}$ and that means that a solution for $f(x)\equiv 0 \pmod{2^3}$ is the class $6 \pmod{8}$.
And, if we solve the system $x\equiv0 \pmod{5}$, $x\equiv6 \pmod{8}$ the final solution is $x\equiv 30 \pmod{40}$.
Could you please tell me if my answer is completely right?
Thank you.