Finding mod of X^2+1 = 0 to have exactly 4 solutions Find a natural number $m$ that is product of 3 prime numbers, and that the equation $x^2+1 \equiv 0 \text { (mod m)}$ has exactly 4 solutions mod m.

The smallest solution is $m=130=2 \cdot 5\cdot 13$, with roots of $x^2+1 \equiv 0 \pmod m$  being $\{47, 57, 73, 83\}$
