Solve (if possible)the congruence involving polynomial


My work:

Since $15=3\cdot5$, we have

$x^3+4x+8\equiv{0}\pmod{3}$ and $x^3+4x+8\equiv{0}\pmod{5}$

In $\mathbb{Z}_3$,

We have $[0],[1],[2]$

They all dont work

In $\mathbb{Z}_5$,

We have $[0],[1],[2],[3],[4]$

They all dont work

So does it mean I have NO solution?

Thank you!!


1 Answer 1


Modulo $3$, there is a solution: $x\equiv 2\pmod{3}$ does work.

But modulo $5$, there is no solution.

So there is no solution modulo $15$. For if $x^3+4x+8\equiv 0\pmod{15}$, then $x^3+4x+8\equiv 0\pmod{5}$.


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