sjLin
February 27, 2022
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m, n Z, m, n ̸= 0, g|mg|nm, n(greatest common divisor)g = gcd(m, n)
g = (m, n)
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m, n Z, gcd(m, n) = 1, mn(relatively prime).
輾轉(Euclidean algorithm)使
m, n Zm = an + ba, b Z, gcd(m, n) = gcd(n, b).
gcd(m, n) = gcd(n, m mod n)
1