已知M=(2+1)(2^2+1)(2^4+1)(2^8+1) ……(2^16已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^16+1)(2^128+1)
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![已知M=(2+1)(2^2+1)(2^4+1)(2^8+1) ……(2^16已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^16+1)(2^128+1)](/uploads/image/z/15145-25-5.jpg?t=%E5%B7%B2%E7%9F%A5M%3D%282%2B1%29%282%5E2%2B1%29%282%5E4%2B1%29%282%5E8%2B1%29+%E2%80%A6%E2%80%A6%282%5E16%E5%B7%B2%E7%9F%A5M%3D%282%2B1%29%282%5E2%2B1%29%282%5E4%2B1%29%282%5E8%2B1%29%E2%80%A6%E2%80%A6%282%5E16%2B1%29%282%5E128%2B1%29)
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1) ……(2^16已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^16+1)(2^128+1)
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1) ……(2^16
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)
……(2^16+1)(2^128+1)
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1) ……(2^16已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^16+1)(2^128+1)
答:补一个因数1=2-1,值不变,利用平方差公式:
M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1).(2^128+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1).(2^128+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1).(2^128+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1).(2^128+1)
=(2^8-1)(2^8+1)(2^16+1).(2^128+1)
=(2^16-1)(2^16+1).(2^128+1)
=(2^32-1).(2^128+1)
=2^256-1
令m=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)
两边同时乘(2-1)得:
(2-1)m=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)
m=(2^2-1)(2^2+1)......(2^128+1)
=......
=(2^128-1)(2^128+1)
=2^256-1
=(2^4)^64-1
=16^64-1
x=64
M=(2+1)(2^2+1)(2^4+1)(2^8+1)
……(2^64+1)(2^128+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
……(2^64+1)(2^128+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
……(2^64+1)(2^128+1)
=(2^4-1)(2^4+1)(2^8+1)
……(2^64+1)(2^128+1)
=2^256-1
供你参考。帮助别人快乐自己!
先乘以(2-1),得到(2^2-1)(2^2+1)(2^4+1)(2^8+1)
……(2^16+1)(2^128+1),根据(x-1)(x+1)=x^2-1,可以得到2^256-1