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For positive integers m and n. Let $F_{n}={2^{2^{n}}}+1$ and $G_{m}={2^{2^{m}}}-1$. Which of the following statements are true?

(a)$F_{n}$ divides $G_{m}$ whenever m>n

(b)$gcd(F_{n},G_{m})=1$ whenver m$\neq$n

(c))$gcd(F_{n},F_{m})=1$ whenver m$\neq$n

(d)$G_{m}$ divides $F_{n}$ whenever m < n

My try

I only able to solve it by plugging values for m and n

Take $m=3;n=2$ then option b is wrong

Take $m=2;n=3$ then optio d is wrong

Therefore a and c are correct options.

But how can I prove a and c options. Anyone can give hint?

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For problem $a$ we first prove that $G_n$ divides $G_{n+1}$, this is easy $2^{2^n}-1|(2^{2^n}+1)(2^{2^n}-1)=G_{n+1}$.

So now take $n<m$.

Notice $F_n|G_{n+1}$ because $2^{2^n}-1|(2^{2^n}-1)(2^{2^n}+1)=G_{n+1}$.

We also have $G_{n+1}|G_m$.


For problem $c$ we prove that $\prod\limits_{i=1}^n F_n=F_{n+1}-2$.

This is because $\prod\limits_{i=1}^n (2^{2^n}+1)=1+2+2^2+2^3+\dots+ 2^{2^n-1}=2^{2^n}-1=F_{n+1}-1$

From here take $n<m$ and notice $F_m-2=F_1F_2\dots F_n\dots F_m-1$. Since $F_n$ is odd and it divides $F_{m}-2$ we conclude $F_n$ and $F_m$ are coprime.

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  • $\begingroup$ What about option a $\endgroup$ – Girish Kumar Chandora Nov 3 '17 at 15:44
  • $\begingroup$ sorry i will not repeat this thing next time $\endgroup$ – Girish Kumar Chandora Nov 3 '17 at 15:51
  • $\begingroup$ no problem, I just needed my morning coffee, have a great day $\endgroup$ – Jorge Fernández Hidalgo Nov 3 '17 at 15:57

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