Editing Quadratic reciprocity (section)

===Jacobi symbol===
{{main|Jacobi symbol}}
The [[Jacobi symbol]] is a generalization of the Legendre symbol; the main difference is that the bottom number has to be positive and odd, but does not have to be prime. If it is prime, the two symbols agree. It obeys the same rules of manipulation as the Legendre symbol. In particular

:<math>\begin{align}
\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}} &= \begin{cases} 1 & n \equiv 1 \bmod{4}\\ -1 & n \equiv 3 \bmod{4}\end{cases} \\
\left( \frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}} &= \begin{cases} 1 & n \equiv 1, 7 \bmod{8}\\ -1 & n \equiv 3, 5\bmod{8}\end{cases} \\
\left( \frac{-2}{n}\right) = (-1)^{\frac{n^2+4n-5}{8}} &= \begin{cases} 1 & n \equiv 1, 3 \bmod{8}\\ -1 & n \equiv 5, 7\bmod{8}\end{cases}
\end{align}</math>

and if both numbers are positive and odd (this is sometimes called "Jacobi's reciprocity law"):

:<math> \left(\frac{m}{n}\right) = (-1)^{\frac{(m-1)(n-1)}{4}}\left(\frac{n}{m}\right).</math>

However, if the Jacobi symbol is 1 but the denominator is not a prime, it does not necessarily follow that the numerator is a quadratic residue of the denominator. Gauss's cases 9) - 14) above can be expressed in terms of Jacobi symbols:

:<math>\left (\frac{M}{p}\right) = (-1)^{\frac{(p-1)(M-1)}{4}} \left(\frac{p}{M}\right ),</math>

and since ''p'' is prime the left hand side is a Legendre symbol, and we know whether ''M'' is a residue modulo ''p'' or not.

The formulas listed in the preceding section are true for Jacobi symbols as long as the symbols are defined. Euler's formula may be written

:<math>\left(\frac{a}{m}\right) =\left(\frac{a}{m \pm 4an}\right), \qquad n \in \Z, m\pm4an>0. </math>

'''Example.'''

:<math>\left (\frac{2}{7} \right ) = \left (\frac{2}{15} \right ) = \left (\frac{2}{23} \right ) = \left (\frac{2}{31} \right ) = \cdots = 1.</math>

2 is a residue modulo the primes 7, 23 and 31:

:<math>3^2 \equiv 2 \bmod{7}, \quad 5^2 \equiv 2 \bmod{23}, \quad 8^2 \equiv 2 \bmod{31}.</math>

But 2 is not a quadratic residue modulo 5, so it can't be one modulo 15. This is related to the problem Legendre had: if <math>\left (\tfrac{a}{m} \right) = -1,</math> then ''a'' is a non-residue modulo every prime in the arithmetic progression ''m'' + 4''a'', ''m'' + 8''a'', ..., if there ''are'' any primes in this series, but that wasn't proved until decades after Legendre.<ref>By [[Peter Gustav Lejeune Dirichlet]] in 1837</ref>

Eisenstein's formula requires relative primality conditions (which are true if the numbers are prime)

:Let <math>a, b, a', b'</math> be positive odd integers such that:
::<math>\begin{align}
\gcd &(a,b) =\gcd(a',b')= 1 \\
&a \equiv a' \bmod{4} \\
&b \equiv b' \bmod{4}
\end{align}</math>
:Then
::<math> \left(\frac{a}{b}\right) \left(\frac{b}{a}\right) =\left(\frac{a'}{b'}\right) \left(\frac{b'}{a'}\right).</math>