四个连续整数的积加上1是一个整数的平方,[证明]

设四个整数是a,a+1,a+2,a+3
则a(a+1)(a+2)(a+3)+1
=[a(a+3)][(a+1)(a+2)]+1
=(a^2+3a)[(a^2+3a)+2]+1
=(a^2+3a)^2+2(a^2+3a)+1
=(a^2+3a+1)^2

(a-1)*a*(a+1)*(a+2)+1=a*(a^2-1)(a+2)+1=a*(a^3+2a^2-a-2)+1=a^2(a^2+2a-1)-2a+1=a^2*(a+1)^2-2a^2-2a+1=[a(a+1)]^2-2[a(a+1)]+1=[a(a+1)+1]^2

a是任意整数

x(x+1)(x+2)(x+3)+1=x*x*x*x+6*x*x*x+11*x*x+6*x+1=(x*x+3x+1)*(x*x+3x+1) 证明完毕