The congruence $\,x^2+1\equiv 0\pmod{99}$ is unsolvable

Question

$x^2+1\equiv 0 \mod 99$

I rearranged the congruence to get $x^2\equiv -1 \mod 99$.

We have an isomorphism $\ \mathbb{Z}/99\mathbb{Z}\to\mathbb{Z}/ 3 \mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/11\mathbb{Z}$.

Solving modulo $3$ first, $\ x^2\equiv -1 \mod 3$.

We find that there are no solutions. Likewise for $x^2\equiv -1\mod 11$.

Therefore $x^2+1\equiv 0 \mod 99$ has no solutions.

Is this correct?

Answer 1

Hints:

$$x^2=-1\pmod{99}\implies x^2=-1\pmod{11}$$

But $\;11\neq1\pmod 4\;$ ...

Answer 2

It is correct, but redundant. You are already done when you show it is unsolvable mod $3.\,$ Indeed, if it is solvable mod $99$ then $\, 3\mid 99\mid x^2+1\,\Rightarrow\, x^2\equiv -1\equiv 2\pmod 3,\,$ which contradicts $\,x\equiv 0,1,\,$ or $\,2\Rightarrow x^2\equiv 0\,$ or $\,1$. There is no need to go further and show it's unsolvable mod $11.$

More generally a quadratic is unsolvable $\bmod n$ if its discriminant is nonsquare $\!\bmod n.\,$ Here $\,x^2+1\equiv 0\,$ has discriminant $\,b^2-4ac = 0^2\!-4(1)1\equiv -4\equiv 2\equiv$ nonsquare $\!\bmod 3\,$ as above.

Further, $\,\Bbb Z/9 \cong \Bbb Z/3 \times \Bbb Z/3\,$ is false, since the latter satisfies $\,3x = 0\,$ but the former does not.