Vtyjkyuk

I am trying to solve $x^2-1=0mod 24$. First I use a theorem which states that $mathbbZ/n simeq mathbbZ/p_1^r_1 times cdots times mathbbZ/p_k^r_k$, where $p_1^r_1cdots p_k^r_k$ is the prime factorization of $n$. Then since $24 = 2^3cdot 3$, we can solve two congruence equations:
$$
x_1 := x in mathbbZ/8 : x^2-1 = 0 mod 2^3 qquad x_2:=xinmathbbZ/3 : x^2-1 = 0 mod 3
$$
Its easy enough to show that $x_1 = 1,3,5,7$, and $x_2 =1,2$. But then the next step is to combine them together to get all of the solutions modulo $24$. This is where I am getting confused because the notes say that the solution is: $x = 1,5,7,11,13,17,19,23$, but I don't see where all of these extra numbers are coming from, or how the above theorem is being applied to get these extra solutions. I know that the theorem implies that the number of solutions in $mathbbZ/24$ will be equal to $lvert x_1rvertlvert x_2rvert=8$, but where the specific numbers come from I have no idea. Any advice or suggested readings would be appreciated!

There are no extra numbers there. The numbers that you mentioned are the numbers from the set $0,1,2,ldots,23$ wich are simultaneously congruent with $1$, $3$, $5$, or $7$ modulo $8(=2^3)$ and congruent with $1$ or $2$ modulo $3$. Take $19$, for instance, and note that $19equiv3pmod8$ and that $19equiv1pmod3$.