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在《圆和二次方程》一书中,给出了任何一组勾股数组a、b、c都可由公式a=m~2-n~2,b=2m~n,c=m~2+n~2表示(这里m、n-奇-偶,m>n,m、n均为自然数),同时指出“abc一定能被60整除”,因为它的证明“已经超出你们的知识范围,这里就不谈了”。为此,笔者给出一种浅显的证明。下面先证两个引理。引理1。任何自然数p若不能被3整除,则p~2-1能被3整除。证明:因为任何不能被3整除的自然数p均可表示勾:p=3k±1(这里k为自然数)而p~2=(3k±1)~2=9k~2±6k+1=3(3k~2±2k)+1,所以p~2-1能被3整除。引理2.任何自然数q若不能被5整除,则q~4-1能被5整除。证明:因为任何不能被5整除的自然数q可表示为q=5l±1,或q=5l±2 (这里l为自然数) 而当q=5l±1时,q-1或q+1能被5整除;当q=5l±2时,q~2=(5l±2)~2
In the book “Circular and Quadratic Equations”, it is shown that any set of pygadian arrays a, b, and c can be represented by the formula a=m~2-n~2, b=2m~n, c=m~2 +n~2 means (where m, n-odd-even, m>n, m, and n are all natural numbers). It also states that “abc must be divisible by 60,” because its proof “has exceeded your knowledge.” I will not talk about it here." For this reason, the author gives a simple proof. Here are two lemmas. Lemma 1. If any natural number p is not divisible by 3, then p~2-1 can be divisible by three. Proof: Because any natural number p that cannot be divisible by 3 can express the hook: p=3k±1 (where k is a natural number) and p~2=(3k±1)~2=9k~2±6k+1=3 ( 3k~2±2k)+1, so p~2-1 can be divisible by three. Lemma 2. If any natural number q is not divisible by 5, q~4-1 can be divisible by 5. Proof: Since any natural number q that is not divisible by 5 can be expressed as q=5l±1, or q=5l±2 (where l is a natural number) and when q=5l±1, q-1 or q+1 can be Divisible by 5; when q=5l±2, q~2=(5l±2)~2