Solve for $m$: $m^5 + 7m + 8 \equiv 0 \pmod 3$

Question

I got about this far:

$$m^5 + 7m \equiv -8 \equiv 1 \pmod 3$$

I'd have a much easier time solving a linear equation, but I have no clue about this one.

$$m(m^4 + 7) \equiv 1 \pmod 3$$

m itself is not divisible by three, but we already knew that... Any thoughts?

Answer 1

You are working in the space $\mathbb{Z}/3\mathbb{Z}=\{0,1,2\}$ so there are only three elements to check.

For $m\equiv 0$, $m^{5}+7m+8=8\equiv 2\mod 3$, so $0$ is not a solution.

For $m\equiv 1$, $m^{5}+7m+8=16\equiv 1\mod 3$, so $1$ is not a solution.

For $m\equiv 2$, $m^{5}+7m+8=54\equiv 0\mod 3$, so $2$ is a solution.

So $2$ is the only solution.

Answer 2

$$m^3\equiv m\pmod3$$

$$m^5\equiv m^3\equiv m\pmod3$$

So, we need $3|8(m+1)\iff m\equiv-1\equiv2\pmod3$ as $(8,3)=1$

Answer 3

Take the coefficients mod $3$: $$m^5+7m+8\equiv m^5 + m+2\equiv 0\pmod 3$$ Apply the Little Theoreom of Fermat, which is $m^p\equiv m \pmod{p}$ for a prime $p$ and any $m$: $$m\cdot m^2 + m+2\equiv m + m+2 \equiv 2m+2 \equiv 0\pmod 3$$ Therefore: $$m \equiv -1 \equiv 2\pmod 3$$

Answer 4

Alternatively: $$m^5+7m+8\equiv m^5-m+8m+8\equiv m(m-1)(m+1)(m^2+1)+8(m+1)\equiv \\8(m+1)\equiv 0\pmod{3}.$$ Here is the linear congruence equation for you.