试求(2-1)(2+1)(2^2+1)......(2^44+1)的个位数

利用平方差公式(a-b)(a+b)=a^2-b^2

(2-1)(2+1)(2^2+1)......(2^44+1)

=(2^2-1)(2^2+1)......(2^44+1)

=(2^4-1)(2^4+1)......(2^44+1)

=......

=(2^44-1)(2^44+1)

=2^88-1

2^1个位数是2
2^2个位数是4
2^3个位数是8
2^4个位数是6
2^5个位数又是2

四个一个循环
又2^88是4的倍数
2^88个位数是6 2^88-1个位数是5

所以(2-1)(2+1)(2^2+1)......(2^44+1)的个位数是5

(2-1)(2+1)(2^2+1)......(2^44+1)
=(2^2-1)(2^2+1)......(2^44+1)
=2^45-1
2^5个位是2
2^45=(2^5)^9=2^9=2*2^4=2^5=2
所以个位是2-1=1