四个连续正整数的积比一个完全平方数少1，为什么？

设a是一个正整数，则
a(a+1)(a+2)(a+3)
=[a(a+3)][(a+1)(a+2)]
=(a²+3a)(a²+3a+2)
=(a²+3a)²+2(a²+3a)+1-1
=[(a²+3a)+1]²-1
=(a²+3a+1)²-1
所以四个连续正整数的乘积比一个完全平方数少1

设这四个数是
a-1.5,a-0.5,a+0.5,a+1.5
(a-1.5)(a-0.5)(a+0.5)(a+1.5)
=[(a-1.5)(a+1.5)][(a-0.5)(a+0.5)]
=[a^2-2.25][a^2-0.25]
设 b=a^2-1.25
那么 (a-1.5)(a-0.5)(a+0.5)(a+1.5)
=[b-1][b+1]
=b^2-1