Editing Quadratic reciprocity (section)

==Statement of the theorem==

'''Quadratic Reciprocity (Gauss's statement).''' If <math>q \equiv 1 \bmod{4}</math>, then the congruence <math>x^2 \equiv p \bmod{q}</math> is solvable if and only if <math>x^2 \equiv q \bmod{p}</math> is solvable. If <math>q \equiv 3 \bmod{4}</math> and <math>p \equiv 3 \bmod{4}</math>, then the congruence <math>x^2 \equiv p \bmod{q}</math> is solvable if and only if <math>x^2 \equiv -q \bmod{p}</math> is solvable.

'''Quadratic Reciprocity (combined statement).''' Define <math>q^* = (-1)^{\frac{q-1}{2}}q</math>. Then the congruence <math>x^2 \equiv p \bmod{q}</math> is solvable if and only if <math>x^2 \equiv q^* \bmod{p}</math> is solvable.

'''Quadratic Reciprocity (Legendre's statement).''' If ''p'' or ''q'' are congruent to 1 modulo 4, then: <math>x^2 \equiv q \bmod{p}</math> is solvable if and only if <math>x^2 \equiv p \bmod{q}</math> is solvable. If ''p'' and ''q'' are congruent to 3 modulo 4, then: <math>x^2 \equiv q \bmod{p}</math> is solvable if and only if <math>x^2 \equiv p \bmod{q}</math> is not solvable.

The last is immediately equivalent to the modern form stated in the introduction above. It is a simple exercise to prove that Legendre's and Gauss's statements are equivalent – it requires no more than the first supplement and the facts about multiplying residues and nonresidues.