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

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1. frx

@myininaya

2. frx

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

3. frx

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

4. myininaya

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....

5. frx

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

6. myininaya

Yeah. lol. I think you are right.

7. frx

Woho! :)

8. frx

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

9. myininaya

The induction part confuses you?

10. myininaya

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

11. frx

Wonderful!

12. frx

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?

13. myininaya

Yes :)

14. myininaya

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

15. myininaya

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

16. frx

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

17. frx

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

18. myininaya

$=2^{2^{k+1}}+1=2^{2^k2^1}+1=2^{2^k 2}+1=(2^{2^k})^2+1$

19. myininaya

They added in a zero.

20. frx

Why did the add the zero for?

21. myininaya

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

22. myininaya

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

23. myininaya

They were setting up for the induction part

24. myininaya

$=(2^{2^k}+1)(2^{2^k}+1)-2^{2^k}$ Ignore that 1oops

25. myininaya

the one in expression i wrote before this one

26. myininaya

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

27. frx

Ok I think I get the whole induction proof now.

28. frx

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

29. myininaya

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

30. frx

* given that m<n

31. myininaya

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

32. frx

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

33. myininaya

no Fn is not equal to 2

34. frx

* F_n is replaces by two *

35. frx

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

36. myininaya

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

37. frx

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

38. frx

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

39. myininaya

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?

40. myininaya

Implies d=1

41. frx

Not at all, that made perfect sense actually

42. frx

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!

43. myininaya

I try.

44. frx

Thank you for your patience, have a great evening!

45. myininaya

You too. :)

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