3
$\begingroup$

Solve (if possible)the congruence involving polynomial

$x^3+4x+8\equiv{0}\pmod{15}$


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!!

$\endgroup$

1 Answer 1

5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.