推 adcores5 :謝謝 07/06 09:25
※ 引述《adcores5 (ok)》之銘言：
: For n>=1,show that gcd(Fn,n)=1
: Fn是費馬數
: Hint:any prime divisor p of Fn (n>=2) is of the form p=k*2^(n+2)+1
: 希望有人可以幫忙，謝謝了!!!
If Fn=2^2^n+1 has a prime factor p, then 2^2^n≡-1(mod p).
=> order of 2 in Z/pZ is 2^{n+1} => 2^{n+1}|φ(p)=p-1 => p=k*2^{n+1}+1>n
=> gcd{Fn,n}=1
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