Here's the question you clicked on:
frx
Proove that two Fermat numbers always are relatively prime. I have the lead that F_m | (F_n - 2) for m<n.
I've tried to understand the solution here: https://www.artofproblemsolving.com/Wiki/index.php/Fermat_number but don't really understand it.
I also have this solution but don't understand it neither
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....
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
Yeah. lol. I think you are right.
But who do they get the recursion formula from \[F _{n}=2^{2^{n}}+1\]
The induction part confuses you?
\[F_0=2^{2^0}+1=2^1+1=2+1=3 \] \[F_1=2^{2^1}+1=2^2+1=4+1=5=3+2 \] \[F_2=2^{2^2}+1=2^4+1=16+1=17=15+2 =3(5)+2=F_0 \cdot F_1 +2 \]
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?
They are proving it :) Proving that will prove gcd(Fm,Fn)=1
If you need help with the algebra, I can do that. Right now, I'm multitasking though...so I will help slowly.
So there are two things I don't understand the first is:\[F _{k+1}=2^{2^{k+1}}+1\] \[=\left( \left( 2^{2^{k}} \right)^{2} +2*2^{2^{k}}+1 \right) - 2*2^{2^{k}}\]
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.
So seeing that \[F _{k+1}=F _{0}*F _{1}...*F _{k-1}*F _{k}+2\] How do I know that \[GCD(F _{m},F _{n})=GCD(F _{m},2)\]
Please ask any questions if you have them. I like this proof.
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
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) ?
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?
Not at all, that made perfect sense actually
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!
Thank you for your patience, have a great evening!