Question

Proove that two Fermat numbers always are relatively prime. I have the lead that F_m | (F_n - 2) for m 13 years ago

Answer 1

@myininaya

Answer 2

I've tried to understand the solution here: https://www.artofproblemsolving.com/Wiki/index.php/Fermat_number but don't really understand it.

Answer 3

I also have this solution but don't understand it neither

Answer 4

Hey, so from that link the only thing I don't understand so far is why all of that shows that they are relatively prime. I do understand everything else though....
So still thinking as to why it shows they are....

Answer 5

I thinks it has to do with the fact that every Fermat number is odd so can not be divided by two and hence the only gcd is 1

Answer 6

Yeah. lol. I think you are right.

Answer 7

Woho! :)

Answer 8

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

Answer 9

The induction part confuses you?

Answer 10

\[F_0=2^{2^0}+1=2^1+1=2+1=3 \]
\[F_1=2^{2^1}+1=2^2+1=4+1=5=3+2 \]
\[F_2=2^{2^2}+1=2^4+1=16+1=17=15+2 =3(5)+2=F_0 \cdot F_1 +2 \]

Answer 11

Wonderful!

Answer 12

So when one have the recursion formula \[F _{n+1}=F _{0}*F _{1}*F _{2}*...*F _{n}+2\]
The solution is to show that the recursion formula is correct using induction, right?

Answer 13

Yes :)

Answer 14

They are proving it :)

Proving that will prove gcd(Fm,Fn)=1

Answer 15

If you need help with the algebra, I can do that.
Right now, I'm multitasking though...so I will help slowly.

Answer 16

So there are two things I don't understand the first is:\[F _{k+1}=2^{2^{k+1}}+1\]
\[=\left( \left( 2^{2^{k}} \right)^{2} +2*2^{2^{k}}+1 \right) - 2*2^{2^{k}}\]

Answer 17

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

Answer 18

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

Answer 19

They added in a zero.

Answer 20

Why did the add the zero for?

Answer 21

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

Answer 22

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

Answer 23

They were setting up for the induction part

Answer 24

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

Answer 25

the one in expression i wrote before this one

Answer 26

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

Answer 27

Ok I think I get the whole induction proof now.

Answer 28

So seeing that \[F _{k+1}=F _{0}*F _{1}...*F _{k-1}*F _{k}+2\]
How do I know that \[GCD(F _{m},F _{n})=GCD(F _{m},2)\]

Answer 29

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

Answer 30

* given that m<n

Answer 31

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

Answer 32

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

Answer 33

no Fn is not equal to 2

Answer 34

* F_n is replaces by two *

Answer 35

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

Answer 36

gcd(Fm,Fn)
=
gcd(Fm,F0F1...F(n-1)+2)

What about this?

Answer 37

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

Answer 38

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

Answer 39

Well F_(n+1) equals
F0F1....Fn+2

----
Anyways,
Fm|(Fn-2)
=>for some integer k we have k*Fm=Fn-2
=> k*Fm+2=Fn

So gcd(Fm,Fn)
= gcd(Fm, k*Fm+2)


Let gcd(Fm, k*Fm+2)=d
=> for some integers a and b we have
aFm+b(k*Fm+2)=d
Fm(a+bk)+2b=d

But a+bk and b are just integers

We know gcd(Fm,2)=1 since Fm is odd and isn't divisible by 2.
This implies for some integers s and t we have sFm+2t=1
where s above is a+bk and t is b from above.

Did this make it more confusing?

Answer 40

Implies d=1

Answer 41

Not at all, that made perfect sense actually

Answer 42

Will have to put it all together and solve it 2-3 times just to make sure it stays in my brain but you're given me the broad understanding, can't thank you enough!

Answer 43

I try.

Answer 44

Thank you for your patience, have a great evening!

Answer 45

You too. :)