Determine the number of distinct solutions of $(x^2-1)(x^2+1)\equiv 0 \pmod {4\cdot31^3}$

Question

I want to determine the number of distinct solutions of $(x^2-1)(x^2+1)\equiv 0 \pmod{4\times 31^3}$,if I call LHS as $f(x)$,then $f(x)\equiv 0 \pmod{4}$ and $f(x)\equiv 0 \pmod{31^3}$.Now I do not know how to proceed.Can someone help me?

Answer 1

$(x^2-1)(x^2+1)\equiv 0\pmod{4\times 31^3}\implies (x^2-1)(x^2+1)\equiv 0\pmod{4}$ and $(x^2-1)(x^2+1)\equiv 0 \pmod{31^3}$

For the first equation, note that $x$ cannot be even,and for $x$ odd,we have $x^2\equiv 1 \pmod{4}$ i.e. $(x^2-1)\equiv 0\pmod{4}$

and for the second equation,we get $x^2-1\equiv 0\pmod{31^3}$ or $x^2+1\equiv 0\pmod{31^3}$

But note that $-1\in Q_{31^3}$ iff $-1\in Q_{31}$ but $31\not\equiv 1\pmod{4}$,so $(x^2+1)\not \equiv 0 \pmod{31^3}$

So finally we get $x^2\equiv 1\pmod{4}$ and $x^2\equiv 1 \pmod{31^3}$

The first equation has two solutions and the second has $\frac{|U_{31^3}|}{|Q_{31^3}|}=2$ solutions.

By chinese remainder theorem,the simultaneous equations has $4$ solutions.