By using the Law of Quadratic Reciprocity find the first primitive root of the prime modulo 223. By using this primitive root, find the square root of $32 \pmod {223}$.
If $r$ is a primitive root of 223 the smallest $j$ such that $r^j \equiv 1 \pmod {223}$ is $j=\phi(223)=222$. We have
$$ r^{222} = 1 \pmod {223} $$
The factors of 222 are $1, 2, 3, 6, 37, 74, 111$, and are the only ones we need to test against candidate values for $r$.
However, recognising that 2, 3, 6 and 37 are factors of 74 and 111, we should test the larger 74 and 111 first to more efficiently determine whether a candidate value of $r$ is primitive or not. That is, if $r^{74} \equiv 1 \pmod {223}$ or $r^{111} \equiv 1 \pmod {223}$ then we don't need to test the smaller factors.
Let's try $r=2$. We want to evaluate $2^{111} \pmod {223}$. By Euler's Criterion $2^{\frac{223-1}{2}} \equiv 2^{111} \equiv 1 \pmod {223}$ if an only if $(\frac{2}{223})=1$. By Proposition (7.15) we have $ (\frac{2}{223}) = 1 $. And so $2^{111}\equiv 1 \pmod {223}$.
This means 2 is not a primitive root of 223.
Let's try $r=3$. We want to evaluate $3^{111} \pmod {223}$. By Euler's Criterion $3^{111} \equiv 1 \pmod {223}$ if and only if $(\frac{3}{223}) =1$. By the result of Exercise (7.3).3(i) we have $(\frac{3}{223}) = -1$. And so $3^{111} \not \equiv 1 \pmod {223}$.
Let's try $3^74$,
$$$ 3^{74} \equiv (3^2)^{37} \equiv (9^6)^6 \times 9 \equiv (32)^6 \times 9 \equiv 183 \pmod {223} $
And so $3^{37} \not \equiv 1 \pmod {223}$.
The following table shows the calculations for smaller factors of 222.
| j | 3^j | 3^j mod 223 |
| 1 | 3 | 3 |
| 2 | 9 | 9 |
| 3 | 27 | 27 |
| 6 | 729 | 60 |
Since none of these factors results in a congruence of 1 modulo 223, we conclude that 3 is indeed a primitive factor of 223.
We need to find $x$ such that
$$ x^2 \equiv 32 \pmod {223} $$
The following table of indices may help.
| n | a = 3^n | a = 3^n mod 223 |
| 1 | 3 | 3 |
| 2 | 9 | 9 |
| 3 | 27 | 27 |
| 4 | 81 | 81 |
| 5 | 243 | 20 |
| 6 | 729 | 60 |
| 7 | 2187 | 180 |
| 8 | 6561 | 94 |
| 9 | 19683 | 59 |
| 10 | 59049 | 177 |
| 11 | 177147 | 85 |
| 12 | 531441 | 32 |
| 13 | 1594323 | 96 |
We can read off that $3^{12} \equiv 32 \pmod {223}$. This means
$$ 3^{12} \equiv (\pm 3^6)^2 \equiv (\pm 60)^2 \equiv 32 \pmod {223} $$
And so the square roots of 32 are $60 \pmod {223}$ and $163 \pmod {223}$.
(Note we didn't use the fact that 3 is a primitive root of 223 to calculate the square roots more algebraically).