Some time ago I have solved an exercise and now, re-reading it, I don't understand a step. I ask your help in that. I will take some results for granted, although in the original exercise they were proven.
Follows the exercise:
Determine the set of odd primes p for which 23 is a quadratic residue.
Solution
We need to find all $p$ such that $\left(23\mid p\right)=1$.
By the Law of Quadratic Reciprocity, it is possible to write $\left(23\mid p\right)=\left(-1\right)^k\left(p\mid23\right)$ where $k=\left(23-1\right)\left(p-1\right)/4$. Evidently, $k$ is even unless $p\equiv3\mod4$. In short, to solve this problem, we need to know the value of $p\mod23$ and $p\mod4$, so we will work modulus 92. Since $\varphi\left(92\right)=44$, there are only 44 cases to consider.
It is possible to prove that the set of QR modulus 23 is $S:=\{1,2,3,4,6,8,9,12,13,16,18\}$ and the set of NQR modulus 23 is $T:=\{5,7,10,11,14,15,17,19,20,21,22\}$.
Now the part I don't understand anymore, and for which I would require some help:
Denote with U be the set of multiples of elements of S congruent to 1 modulo 4 not exceeding 92, and with V the set of multiples of elements of T congruent to 3 modulo 4 not exceeding 92. By construction, these sets contain only odd elements; in fact:
$U = \{1,9,13,25,29,41,49,73,77,81,85\}$
and
$V = \{7,11,15,19,43,51,63,67,79,83,91\}$
Now my question is: how did I build these two sets? I can't really remember – and I tested all the possibilities without success.
The remaining part of the exercise is clear to me:
$(23\mid p)=1$ iff $p\equiv r\mod92$, where r belongs to the union of $U$ and $V$.
Best Answer
For the set U, expand the set of QR to be modulo 92 (adding 23 zero, once, twice and three times) giving {1,2,3,4,6,8,9,12,13,16,18,24,25,26,27,29,31,32,35,36,39,41,47,48,49,50,52,54,55,58,59,62,64,70,71,72,73,75,77,78,81,82,85,87} Then as QR requires p=1(mod 4) in order for the multiplication to give 1 x 1 = 1, we "filter out" those values which do not have the correct modulo 4. This filtering gives the set U
Similarly for V, you filter out those =3 mod4 as they require multiplication -1 x -1 = 1