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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?

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  • $\begingroup$ $\large \bmod 3\!:\ \, m\not\equiv 0\,\Rightarrow\, m\equiv \pm1\,\Rightarrow\,\color{#C00}{m^4\equiv 1},\ $ so $\large \ 1\equiv (\color{#c00}{m^4}\!+\!7)m\equiv -m\ \ $ $\endgroup$
    – Bill Dubuque
    Dec 17, 2018 at 18:54
  • $\begingroup$ The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though. $\endgroup$
    – Jyrki Lahtonen
    Dec 19, 2018 at 10:30

4 Answers 4

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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.

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  • $\begingroup$ Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time? $\endgroup$
    – Minah
    Dec 17, 2018 at 18:18
  • $\begingroup$ For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues. $\endgroup$
    – pwerth
    Dec 17, 2018 at 18:23
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$$m^3\equiv m\pmod3$$

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

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

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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: $$m \equiv -1 \equiv 2\pmod 3$$

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  • $\begingroup$ This is essentially a dupe of lab's answer. $\endgroup$
    – Bill Dubuque
    Dec 17, 2018 at 18:40
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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.

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