Editing Quadratic reciprocity (section)

==Supplements to quadratic reciprocity==

The supplements provide solutions to specific cases of quadratic reciprocity. They are often quoted as partial results, without having to resort to the complete theorem.

===''q'' = ±1 and the first supplement===

Trivially 1 is a quadratic residue for all primes. The question becomes more interesting for −1. Examining the table, we find −1 in rows 5, 13, 17, 29, 37, and 41 but not in rows 3, 7, 11, 19, 23, 31, 43 or 47. The former set of primes are all congruent to 1 modulo 4, and the latter are congruent to 3 modulo 4.

:'''First Supplement to Quadratic Reciprocity.''' The congruence <math>x^2 \equiv -1 \bmod{p}</math> is solvable if and only if <math>p</math> is congruent to 1 modulo 4.

===''q'' = ±2 and the second supplement===
Examining the table, we find 2 in rows 7, 17, 23, 31, 41, and 47, but not in rows 3, 5, 11, 13, 19, 29, 37, or 43. The former primes are all ≡ ±1 (mod 8), and the latter are all ≡ ±3 (mod 8). This leads to

:'''Second Supplement to Quadratic Reciprocity.''' The congruence <math>x^2 \equiv 2 \bmod{p}</math> is solvable if and only if <math>p</math> is congruent to ±1 modulo 8.

−2 is in rows 3, 11, 17, 19, 41, 43, but not in rows 5, 7, 13, 23, 29, 31, 37, or 47. The former are ≡ 1 or ≡ 3 (mod 8), and the latter are ≡ 5, 7 (mod 8).

===''q'' = ±3===
3 is in rows 11, 13, 23, 37, and 47, but not in rows 5, 7, 17, 19, 29, 31, 41, or 43. The former are ≡ ±1 (mod 12) and the latter are all ≡ ±5 (mod 12).

−3 is in rows 7, 13, 19, 31, 37, and 43 but not in rows 5, 11, 17, 23, 29, 41, or 47. The former are ≡ 1 (mod 3) and the latter ≡ 2 (mod 3).

Since the only residue (mod 3) is 1, we see that −3 is a quadratic residue modulo every prime which is a residue modulo 3.

===''q'' = ±5===
5 is in rows 11, 19, 29, 31, and 41 but not in rows 3, 7, 13, 17, 23, 37, 43, or 47. The former are ≡ ±1 (mod 5) and the latter are ≡ ±2 (mod 5).

Since the only residues (mod 5) are ±1, we see that 5 is a quadratic residue modulo every prime which is a residue modulo 5.

−5 is in rows 3, 7, 23, 29, 41, 43, and 47 but not in rows 11, 13, 17, 19, 31, or 37. The former are ≡ 1, 3, 7, 9 (mod 20) and the latter are ≡ 11, 13, 17, 19 (mod 20).

===Higher ''q''===
The observations about −3 and 5 continue to hold: −7 is a residue modulo ''p'' if and only if ''p'' is a residue modulo 7, −11 is a residue modulo ''p'' if and only if ''p'' is a residue modulo 11, 13 is a residue (mod ''p'') if and only if ''p'' is a residue modulo 13, etc. The more complicated-looking rules for the quadratic characters of 3 and −5, which depend upon congruences modulo 12 and 20 respectively, are simply the ones for −3 and 5 working with the first supplement.

:'''Example.''' For −5 to be a residue (mod ''p''), either both 5 and −1 have to be residues (mod ''p'') or they both have to be non-residues: i.e., ''p'' ≡ ±1 (mod 5) ''and'' ''p'' ≡ 1 (mod 4) or ''p'' ≡ ±2 (mod 5) ''and'' ''p'' ≡ 3 (mod 4). Using the [[Chinese remainder theorem]] these are equivalent to ''p'' ≡ 1, 9 (mod 20) or ''p'' ≡ 3, 7 (mod 20).

The generalization of the rules for −3 and 5 is Gauss's statement of quadratic reciprocity.