Legendre Symbol¶
\[ \left ( \frac{a}{p} \right ) = \begin{cases} 1 \qquad &\text{if } \exists x \text{ that } x^2 \equiv a \pmod p \text{ and } a \not\equiv 0 \pmod p \\ -1\qquad &\text{if } \forall x \text{ that } x^2 \not\equiv a \pmod p \text{ and } a \not\equiv 0 \pmod p \\ 0\qquad &\text{if } a \equiv 0 \pmod p \end{cases} \]
we can write the Legendre Symbol like (if \(p\) is prime)
\[ \left ( \frac{a}{p} \right ) = a^{ \frac{p-1}{2} } \pmod p \]
Proof¶
we know that \(a^{p-1} \equiv 1 \pmod p\),
\[ (a^{ \frac{p-1}{2}} + 1) \cdot (a^{ \frac{p-1}{2}} - 1) \equiv 0 \pmod p \]
so \(a^{\frac{p-1}{2}}\) will be \(1\) or \(-1\)
if \(a\) is \(p\)’s quadratic residue, then \(\exists x\) that \(x^2 \equiv a \pmod p\), then
\[ a^{\frac{p-1}{2}} \equiv x^{p-1} \equiv 1 \pmod p \]
if \(a^{\frac{p-1}{2}} \equiv 1 \pmod p\), because \(p\) is a prime number, so \(p\)’s primitive root exist. assume \(d\) is \(p\)’s primitive root, then \(\exists j\) that \(1 \le j \le p-1\) let \(a \equiv d^j \pmod p\), so
\[ a^{\frac{p-1}{2}} \equiv d^{j \cdot \frac{p-1}{2}} \equiv 1 \pmod p \]
because \(d\) is \(p\)’s primitive root, so \((p - 1) | j \cdot \frac{p-1}{2}\) \(\Rightarrow\) \(j\) is even \(\Rightarrow\) \(a\) is \(p\)’s quadratic residue
Code¶
def legendre_symbol(a: int, p: int):
"""
- input : `a (int)`, `p (int)`
- output : `ls (int)` , value of (a/p) (legendre symbol)
"""
ls = pow(a, (p - 1) // 2, p)
return -1 if ls == (p - 1) else ls
Jacobi Symbol¶
for any integer \(a\) and any positive odd integer \(n = {p_1}^{\alpha_1}{p_2}^{\alpha_2}...{p_k}^{\alpha_k}\)
\[ \left ( \frac{a}{n} \right ) = \left ( \frac{a}{p_1} \right )^{\alpha_1} \left ( \frac{a}{p_2} \right )^{\alpha_2} ... \left ( \frac{a}{p_k} \right )^{\alpha_k} \]
Properties :
- If \(( \frac{a}{n} ) = -1\) then \(a\) is a quadratic nonresidue modulo \(n\)
- If \(a\) is a quadratic residue modulo \(n\) and \(\operatorname{gcd}(a, n) = 1\), then \(( \frac{a}{n} ) = 1\)
- If \(( \frac{a}{n} ) = 1\) then \(a\) may or may not be a quadratic residue modulo \(n\)