All integers $m$ satisfying $m\equiv n^2 \mod 19$

Question

1. How to find all Integers $m$ satisfying $m\equiv n^2 \mod 19$ for some $n\in Z$ ?

2. For each non-zero element of $Z_{19}$, find its order (in the multiplicative group of this ring).

I don't know how to do both. For the first one, taking all $a$ from $0$ to $19$ we can check some $x$. I got all values of $a$, but don't know what is the general procedure to be followed.

I guess the answers of $a$ should be $a=\left\{0,1,4,5,6,7,9,11,16,17\right\}.$

Answer 1

The multiplicative group of a finite field is cyclic.

The order of $x\in\Bbb Z_{19}^*$ divides $18$ (the order of the group).

Now, there are $\varphi (18)=6$ generators.

By process of elimination, there are $\varphi(3)=2$ elements of order $3$. They are $\{7,12\}$.

Next, there are $\varphi (6)=2$ of order $6$. They are $\{8,11\}$.

There are $\varphi (9)=6$ elements of order $9$. We can take them from the list in part $a)$, since a square can't generate. Those are: $\{4,9,16,6,17,5\}$.

Now we can do the $6$ generators: $\{2,3,10,13,14,15\}$.

The only one left is $\{18\}$, which has order $2$.

Answer 2

1. We can just take the residues of $0^2,1^2,2^2,\cdots, 9^2$, see here:

For an odd prime $p$, prove that the quadratic residues of $p$ are congruent modulo $p$ to the integers

So these are $m=1,4,9,16,6,17,11,7,5$ as quadratic residues modulo $19$. Furthermore $m=0$ is a square.