冪剰余記号
( α p ) n ≡ α N p − 1 n mod p . {\displaystyle \left({\frac {\alpha }{\mathfrak {p}}}\right)_{n}\equiv \alpha ^{\frac {\mathrm {N} {\mathfrak {p}}-1}{n}}{\bmod {\mathfrak {p}}}.}
( α p ) n ( β p ) n = ( α β p ) n {\displaystyle \left({\frac {\alpha }{\mathfrak {p}}}\right)_{n}\left({\frac {\beta }{\mathfrak {p}}}\right)_{n}=\left({\frac {\alpha \beta }{\mathfrak {p}}}\right)_{n}}
α ≡ β mod p ⇒ ( α p ) n = ( β p ) n {\displaystyle \alpha \equiv \beta {\bmod {\mathfrak {p}}}\quad \Rightarrow \quad \left({\frac {\alpha }{\mathfrak {p}}}\right)_{n}=\left({\frac {\beta }{\mathfrak {p}}}\right)_{n}}