Bézout恒等式
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(Bézout / 裴蜀 / 贝祖 / 比舒)
In elementary number theory, Bézout's identity (also called Bézout's lemma) is the following theorem: Bézout's identity — Let a and b be integers with greatest common divisor d Then, there exist integers x and y such that ax + by = d More generally, the integers of the form ax + by are exactly the multiples of d ——wikipedia |
译: 在初等数论中,Bézout恒等式(也称为Bézout引理)是下列引理: Bézout恒等式: 设a和b为具有最大公因数d的整数 存在整数x和y,使得ax+by=d 即ax+by恰好是d的倍数 |
wikipedia上说的很清楚,就不再重复说了
(某一种证法)
有a,b∈Z*
记d == gcd(a,b),对ax + by == d,两边同时除以d,可得(a1)x + (b1)y == 1,其中gcd(a1,b1) == 1
转证(a1)x + (b1)y == 1,由带余除法:
① (a1) == (q1)(b1) + (r1),其中0 < r1 < b1
② (b1) == (q2)(r1) + (r2),其中0 < r2 < r1
③ (r1) == (q3)(r2) + (r3),其中0 < r3 < r2
.....
④ (rn-4) == (qn-2)(rn-3) + (rn-2)
⑤ (rn-3) == (qn-1)(rn-2) + (rn-1)
⑥ (rn-2) == (qn)(rn-1) + (rn)
⑦ (rn-1) == (qn+1)(rn) + 1
故,由⑦和⑥推出(rn-2)An-2 + (rn-1)Bn-1 == 1
再结合⑤推出(rn-3)An-3 + (rn-2)Bn-2 == 1
再结合④推出(rn-4)An-4 + (rn-3)Bn-3 == 1
.....
再结合③推出(r1)A1 + (r2)B2 == 1
再结合②推出(b1)A0 + (r1)B0 == 1
再结合①推出(a1)x + (b1)y == 1
证毕
——bia度百科
- n个整数间
设有a1,a2,a3......an为n个整数,d是它们的最大公约数,那么存在整数x1......xn使得x1*a1 + x2*a2 + ... + xn*an == d
——bia度百科