The following facts, even the reciprocity laws, are straightforward deductions from the definition of the Jacobi symbol and the corresponding properties of the Legendre symbol. The Jacobi symbol is defined only when the upper argument ("numerator") is an integer and the lower argument ("denominator") is a positive odd integer. :1. If
n is (an odd) prime, then the Jacobi symbol is equal to (and written the same as) the corresponding Legendre symbol. :2. If , then \biggl(\frac{a}{n}\biggr) = \left(\frac{b}{n}\right) = \left(\frac{a \pm k \cdot n}{n}\right). :3. \biggl(\frac{a}{n}\biggr) = \begin{cases} \, 0 & \text{if } \gcd(a,n) \ne 1,\\ \pm1 & \text{if } \gcd(a,n) = 1. \end{cases} If either the top or bottom argument is fixed, the Jacobi symbol is a
completely multiplicative function in the remaining argument: :4. \;\left(\frac{ab}{n}\right)\; = \biggl(\frac{a}{n}\biggr)\left(\frac{b}{n}\right),\quad\text{so } \left(\frac{a^2}{n}\right) = \biggl(\frac{a}{n}\biggr)^2 = +1 \text{ or } 0. :5. \biggl(\frac{a}{mn}\biggr)=\biggl(\frac{a}{m}\biggr)\biggl(\frac{a}{n}\biggr),\quad\text{so } \left(\frac{a}{n^2}\right) = \biggl(\frac{a}{n}\biggr)^2 = +1 \text{ or } 0. The
law of quadratic reciprocity: if
m and
n are odd positive
coprime integers, then :6. \biggl(\frac{m}{n}\biggr)\biggl(\frac{n}{m}\biggr) = (-1)^{\tfrac{m-1}{2}\cdot\tfrac{n-1}{2}} = \begin{cases} +1 & \text{if } n \equiv 1 \text{ or } m \equiv 1 \pmod 4,\\ -1 & \text{if } n \equiv m \equiv 3 \pmod 4. \end{cases} and its supplements :7. \;\left(\frac{1}{n}\right)\; = \biggl(\frac{n}{1}\biggr) = +1. :8. \left(\frac{-1}{n}\right) = (-1)^\tfrac{n-1}{2} \,= \begin{cases} +1 & \text{if }n \equiv 1 \pmod 4,\\ -1 & \text{if }n \equiv 3 \pmod 4. \end{cases} :9. \;\left(\frac{2}{n}\right)\; = (-1)^\tfrac{n^2-1}{8} = \begin{cases} +1 & \text{if }n \equiv 1,7 \pmod 8,\\ -1 & \text{if }n \equiv 3,5 \pmod 8. \end{cases} Combining properties 4 and 9 gives: :10. \left(\frac{2a}{n}\right) = \left(\frac{2}{n}\right) \biggl(\frac{a}{n}\biggr) = \begin{cases} \phantom{-}\left(\frac{a}{n}\right) & \text{if }n \equiv 1,7 \pmod 8,\\ - \left(\frac{a}{n}\right) & \text{if }n \equiv 3, 5\pmod 8. \end{cases} Combining properties 2, 6 and 10 gives: :11. If , then \biggl(\frac{a}{n}\biggr) = \biggl(\frac{a}{m}\biggr) = \biggl(\frac{a}{n + 4k\left|a\right|}\biggr). {{cite book |first=Henri |last=Cohen |author-link=Henri Cohen (number theorist) |title=A Course in Computational Algebraic Number Theory |year=1996 |orig-year=1993 |isbn=0-387-55640-0 |page=28 Theorem 1.4.9.(4) :12. \biggl(\frac{a}{n+2\left|a\right|}\biggr) = \biggl(\frac{a}{n}\biggr)(-1)^\tfrac{a(a-1)}{2} = \begin{cases} \phantom{-}\left(\frac{a}{n}\right) & \text{if }a \equiv 0,1 \pmod 4,\\ - \left(\frac{a}{n}\right) & \text{if }a \equiv 2,3 \pmod 4. \end{cases} = 0. a = 0 cases give the sign of (a'/n). • Combining the two sigs, the property is shows to hold for even a. • If a is odd, then 2a'≡1 (mod 4), so (-1/n+2a') = -(-1/n). • If a is odd, then a'≡a+2 (mod 4), so the sign of (a'/n) is the opposite of what we want. • The two signs combine to produce the desired property for odd a. --> Like the Legendre symbol: • If = −1 then
a is a quadratic nonresidue modulo
n. • If
a is a
quadratic residue modulo
n and
gcd(
a,
n) = 1, then = +1. But, unlike the Legendre symbol: • If = +1 then
a may or may not be a quadratic residue modulo
n. • If
a is a quadratic nonresidue modulo
n, then may be −1 or +1. This is because for
a to be a quadratic residue modulo
n, it has to be a quadratic residue modulo
every prime factor of
n. However, the Jacobi symbol equals +1 if, for example,
a is a non-residue modulo exactly two of the prime factors of
n. Although the Jacobi symbol cannot be uniformly interpreted in terms of squares and non-squares, it can be uniformly interpreted as the sign of a permutation by
Zolotarev's lemma. The Jacobi symbol is a
Dirichlet character to the modulus
n. ==Calculating the Jacobi symbol==