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Yeah. lol. I think you are right.

Woho! :)

But who do they get the recursion formula from \[F _{n}=2^{2^{n}}+1\]

The induction part confuses you?

Wonderful!

Yes :)

They are proving it :)
Proving that will prove gcd(Fm,Fn)=1

The second thing is \[\gcd(F _{m},F _{n})=\gcd(F _{m},2)\]

\[=2^{2^{k+1}}+1=2^{2^k2^1}+1=2^{2^k 2}+1=(2^{2^k})^2+1\]

They added in a zero.

Why did the add the zero for?

\[=(2^{2^k}+1)^2+1-2 \cdot 2^{2^k} \]
So they can write that one part as Fk

or F_0F_1F_2...F_(K-1)+2

They were setting up for the induction part

\[=(2^{2^k}+1)(2^{2^k}+1)-2^{2^k}\]
Ignore that 1oops

the one in expression i wrote before this one

We are trying to show F_0F_1....F_k+2

Ok I think I get the whole induction proof now.

Please ask any questions if you have them. I like this proof.

* given that m

Well gcd(odd,2)=1 since odd aren't divisible by 2

Sorry formulated it bad, I mean how do I know that gcd F_n is 2 ?

no Fn is not equal to 2

* F_n is replaces by two *

So where does the replacement of F_n to 2 come from?

gcd(Fm,Fn)
=
gcd(Fm,F0F1...F(n-1)+2)
What about this?

Sorry don't understand, Isn't that equal to f_(n+1)?

Has it to do with the initial lead F_m | (F_n - 2) ?

Implies d=1

Not at all, that made perfect sense actually

I try.

Thank you for your patience, have a great evening!

You too. :)