Solving quadratic modular equation.

Question

Solve the system: $x \equiv 2 \pmod{2}$ and $(x-1)(x+1)(x^2+1) \equiv 0 \pmod{5}$.

I got this from a challenge problem; the answer is $x = 2$.

Well we know that $x = 6, 4, 2$ the possibilities I mean. But, which one exactly?

Answer 1

Because 5 is a prime, $(x-1)(x+1)(x^2+1) \equiv 0 \pmod{5}$ iff $x-1 \equiv 0 \pmod{5}$ or $x+1 \equiv 0 \pmod{5}$ or $x^2+1 \equiv 0 \pmod{5}$.

The first two are trivial and give, $x \equiv \pm 1 \pmod{5}$. The second one gives $x \equiv \pm 2 \pmod{5}$, just check all cases.

Therefore all even numbers $x$ not divisible by 5 will statistify, so 2, 4, 6, 8, 12, 14, 16, 18, 22, 24, ...

Answer 2

$$0\equiv(x-1)(x+1)(x^2+1)\pmod5\equiv x^4-1$$

which holds true if $(x,5)=1$ by Fermat's little theorem

So, $x\equiv1,2,3,4\pmod5$ and $x\equiv2\pmod2\equiv0$

$\implies x\equiv2,4,6,8\pmod{10}$