令p = 2m+1,m∈Z
p^2 = (2m+1)^2 = 4m^2+4m+1 = 2(2m^2+2m)+1
∵m∈Z
∴2m^2+2m∈Z
∴2(2m^2+2m)是偶数
∴p^2 = 2(2m^2+2m)+1为奇数3次方或者更高次方呢？p^3 = (2m+1)^3= (2m)^3+3*(2m)^2+3*(2m)*1+1展开后前边各项中均包含因数2m，除了最后的一个1之外，前面各项之和是各个偶数之和，偶数之和仍为偶数，偶数+1为奇数。无论再多次方，都是这个规律。====展开后前边各项中均包含因数2m，除了最后的一个1之外，前面各项之和是各个偶数之和，偶数之和仍为偶数，偶数+1为奇数。无论在多次方，都是这个规律。怎么证明这句话令p = 2m+1，m∈Zp^2 = (2m+1)^2 = (2m)^2+2*(2m)+1p^3 = (2m+1)^3 = (2m)^3+3*(2m)^2+3*(2m)+1p^4 = (2m+1)^4 = (2m)^4+4*(2m)^3+6*(2m)^2+4*(2m)+1p^5 = (2m+1)^5 = (2m)^5+5*(2m)^4+10*(2m)3+10*(2m)^2+5*(2m)+1......p^n = (2m+1)^n = (2m)^n+ n * (2m)^(n-1)+ n(n-1)/(1*2) * (2m)^(n-2) + n(n-1)(n-2)/(1*2*3) * (2m+1)^(n-3) + n(n-1)(n-2)(n-3)/(1*2*3*4) * (2m+1)^(n-4)+......+ n(n-1)(n-2)(n-3)/(1*2*3*4) * (2m)^4+ n(n-1)(n-2)/(1*2*3) * (2m)^3 + n(n-1)/(1*2) * (2m)^2 + n * (2m)+ 1可以看出，展开式除了最后一项之外，前面各项都包含因子（2m)，即前面各项都是偶数，偶数之和为偶数，偶数+1为奇数。============================================================================另外，用数学归纳法也可以：p为奇数，令p=2m+1，m、n为正整数：首先，n=2时：p^2 = (2m+1)^2 = (2m)^2+2*(2m)+1 = 2*2m(m+1)+1为奇数成立假设n=k（k为正整数）时p^k为奇数成立，令p^n = 2W+1那么当n=k+1时：p^(k+1) = p^k*p = (2W+1)*(2m+1) = 4mW+2W+2m+1 = 2(2mW+W+m)+1为奇数成立得证。