最大(小)公约(倍)数相等的刻画

The depiction of the greatest(smallest)common divisor(multiple)'s equality

利用整数可逆矩阵给出了2组整数的最大公约数与最小公倍数分别对应相等的判别定理,得到主要结果为:设ai,bi∈Z(i=1,…,n,n∈Z+,n≥2),则(1)gcd{a1,…,an}=gcd{b1,…,bn}当且仅当存在n阶整数可逆矩阵P,使得(a1,…,an)P=(b1,…,bn),其中:gcd{c1,…,cn}表示整数c1,…,cn的最大公约数;(2)[a1,…,an]=[b1,…,bn]当且仅当存在n阶整数可逆矩阵Q,适合b1…bn(M1,…,Mn)Q=a1…an(N1,…,Nn),其中:aiMi=a1…an,biNi=b1…bn,且aibi≠0(i=1,…,n).

By using of integral invertible matrix gave discriminant theorem for the corresponding equality of the greatest（smallest）common divisor（multiple）of two groups of integerst.he main conclusion as followl,etai,bi∈ Z（i=1,…,n,n∈Z ＋,n≥ 2）,then（1）gcd{a1,…,an} =gcd{b1,…,bn } if and only if there exists ann×n integral invertible matrixP which satisfy（a1,…,an）P=（b1,…,bn）,wheregcd{ c1,…,cn } is the greatest common divisor of integers c1,…,cn;（2）[a1,…,an ]=[b1,…,bn ] if and only if there exists ann × n integral invertible matrixQ which satisfy b1…bn（M1,…,Mn）Q =a1…an（N1,…,Nn）,whereaiMi=a1…an,biN i=b1…bn,andaibi≠ 0（i=1,…,n）.