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. Statement of the theorem Quadratic Reciprocity (Gauss's statement). If , then the congruence is solvable if and only if is solvable. If and , then the congruence is solvable if and only if is solvable. Quadratic Reciprocity (combined statement). Define . Then the congruence is solvable if and only if is solvable. Quadratic Reciprocity (Legendre's statement). If p or q are congruent to 1 modulo