Discrete math(Solve for congruence)

Question

Is my understanding correct? I am trying to solve the congruence of $x^3-x ≡ 0($mod $3)$ Should I be factoring the left to make it $x(x+1)(x-1)$? Whats the next step after I factor?

Remark

Answer 1

As $3$ is prime if you have $m\cdot v \equiv 0 \pmod 3$ (assuming $m$ and $v$ are integers) then that means $3|m\cdot v$ and, because $3$ is prime that means either $3|m$ or $3|v$ or in other words, either $m \equiv 0 \pmod 3$ or $v\equiv 0 \pmod 3$.

So if $x(x+1)(x-1)\equiv 0\pmod 3$ that means either

1. $x \equiv 0 \pmod 3$
2. $x +1 \equiv 0 \pmod 3$ which means $x\equiv -1 \pmod 3$
3. $x -1\equiv 0 \pmod 3$ which mean $x \equiv 1\pmod 3$.

Notice that those are the only possible for options for $x$ so that means $x^3 - x\equiv 0$ for EVERY possible integer $x$!

Another way of thinking this is to note $x-1, x, x+1$ are $3$ consecutive integers. As every third number is a multiple of $3$ one of these $3$ must be a multiple of $3$. So $(x-1)x(x+1) = x^3 - x$ is divisible by $3$.

Also note: as a result $x^3 - x \equiv 0 \pmod 3$ for all integer $x$ then $x^3 \equiv x \pmod 3$ for all integers $x$.

Indeed if $x\equiv 0 \pmod 3$ then $x^3 \equiv 0 \pmod 3$ and if $x\equiv -1 \pmod 3$ then $x^3 \equiv -1 \pmod 3$ and if $x\equiv 1\pmod 3$ then $x^3 \equiv 1 \pmod 3$.

You might be wondering about $2 \pmod 3$. But $2 \equiv -1 \pmod 3$. And if you don't buy that. If $x \equiv 2 \pmod 3$ then $x^3 \equiv 8 \pmod 3$ but $8 \equiv 2 \pmod 3$. So $x^3 \equiv x \pmod 3$.