Question

若2的80次方+2的n次方+2的100次方是一个完全平方数,则n=?

Answer 1

2^80=(2^40)^2

2^100=(2^50)^2

2^n=[2^(n/2)]^2

(2^40+2^50)^2

=(2^40)^2+2×(2^40)×(2^50)+(2^50)^2

=2^80+2^91+2^100

n=91

(2^(n/2)+2^40)^2

=[2^(n/2)]^2+2×[2^(n/2)]×2^40+(2^40)^2

=2^n+2^(41+(n/2))+2^80

由此得41+(n/2)=100,解得n=118

(2^(n/2)+2^50)^2

=[2^(n/2)]^2+2×[2^(n/2)]×2^50+(2^50)^2

=2^n+2^(51+(n/2))+2^100

由此得51+(n/2)=80,解得n=58

n可以取三个值,91,118,58