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)是偶數(shù)
∴p^2 = 2(2m^2+2m)+1為奇數(shù)3次方或者更高次方呢？p^3 = (2m+1)^3= (2m)^3+3*(2m)^2+3*(2m)*1+1展開后前邊各項中均包含因數(shù)2m，除了最后的一個1之外，前面各項之和是各個偶數(shù)之和，偶數(shù)之和仍為偶數(shù)，偶數(shù)+1為奇數(shù)。無論再多次方，都是這個規(guī)律。====展開后前邊各項中均包含因數(shù)2m，除了最后的一個1之外，前面各項之和是各個偶數(shù)之和，偶數(shù)之和仍為偶數(shù)，偶數(shù)+1為奇數(shù)。無論在多次方，都是這個規(guī)律。怎么證明這句話令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，展開式除了最后一項之外，前面各項都包含因子（2m)，即前面各項都是偶數(shù)，偶數(shù)之和為偶數(shù)，偶數(shù)+1為奇數(shù)。============================================================================另外，用數(shù)學(xué)歸納法也可以：p為奇數(shù)，令p=2m+1，m、n為正整數(shù)：首先，n=2時：p^2 = (2m+1)^2 = (2m)^2+2*(2m)+1 = 2*2m(m+1)+1為奇數(shù)成立假設(shè)n=k（k為正整數(shù)）時p^k為奇數(shù)成立，令p^n = 2W+1那么當n=k+1時：p^(k+1) = p^k*p = (2W+1)*(2m+1) = 4mW+2W+2m+1 = 2(2mW+W+m)+1為奇數(shù)成立得證。