(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1) 已知x^2-y^2=8,x+y=4,求x,y的
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(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1) 已知x^2-y^2=8,x+y=4,求x,y的值
(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3*3*(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3(2^2 -1)*(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3(2^4 -1)*(2^4 +1)(2^8 +1)...(2^32 +1)
...
=1/3(2^64-1)
x+y=4 (1)
x^2-y^2=(x+y)(x-y)=4(x-y)=8
x-y=2 (2)
联立(1)(2)解得x=3,y=1
=1/3*3*(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3(2^2 -1)*(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3(2^4 -1)*(2^4 +1)(2^8 +1)...(2^32 +1)
...
=1/3(2^64-1)
x+y=4 (1)
x^2-y^2=(x+y)(x-y)=4(x-y)=8
x-y=2 (2)
联立(1)(2)解得x=3,y=1
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