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定理若有素数p≥3,对于一切非p整倍数的自然数x,有x~r-1除以p所得的余数都是1。下面给出证明。对于一切非p整倍数的自然数x,都能表示为p_m+1,p_m+2,p_m+3,……,p_m+(p-1),其中m是零或自然数。根据二项式展开定理可知,x_(r-1)除以P所得的余数,一定是1~(,-1),2~(,-1),3~(,-1),……,(p-1)~(,-1)除以p所得的余数。所以,问题归结为证明1~(,-1),2~(,-1),3~(,-1),……(p-1)~(,-1)除以p所得的余数均为1。 1~(,-1)=1,∴1~(,-1)=O·p+1
If the theorem has a prime number p ≥ 3, for all non-p integer multiples of natural number x, the remainder of x~r-1 divided by p is all 1. The proof is given below. For all non-p integer multiples of the natural number x, it can be expressed as p_m+1, p_m+2, p_m+3, ..., p_m+(p-1), where m is zero or a natural number. According to the binomial expansion theorem, the remainder of x_(r-1) divided by P must be 1~(, -1), 2~(, -1), 3~(, -1), ..., (p-1) ~ (, -1) Divided by p. So, the problem boils down to the proof that 1~(, -1), 2~(, -1), 3~(, -1),...(p-1)~(, -1) are all divided by p. Is 1. 1~(,-1)=1,∴1~(,-1)=O·p+1