Distinct Fermat numbers are coprime without recursion

Asked Dec 28 '23 at 21:53

Active Dec 28 '23 at 22:01

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For m>n $$\ gcd(F_m,F_n)=gcd(F_m \cdot F_n -F_m,F_n)=gcd(F_m\cdot (F_n-1),F_n)$$

Since $$F_m \cdot (F_n-1) \ is\ even \ and \ F_n \ is \ odd \gcd(F_m,F_n)=1 $$ I am in introduction to number theory so I excuse possible wrong statements and thank you for your help.

edited Dec 28 '23 at 22:01

Anne Bauval

49,005

asked Dec 28 '23 at 21:53

Arthr

See https://en.wikipedia.org/wiki/Fermat_number#Basic_properties – lhf Dec 28 '23 at 22:04
Please do not repost your prior dupe closed questions. – Bill Dubuque Dec 28 '23 at 23:18
For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Dec 28 '23 at 23:19
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Your argument, if it worked, could be used to show that every two odd numbers are relatively prime. That is obviously wrong. – Thomas Andrews Dec 28 '23 at 23:24
@TonyK $F_n-1$ is not necessarily even. – Anne Bauval Dec 29 '23 at 06:21
@ThomasAndrews Is there any way to fix this? – Arthr Dec 29 '23 at 06:45
@TonyK Yes, thank you, I realized my confusion 5 hours ago and already deleted my comments right away. I think you can delete yours as well. – Anne Bauval Dec 29 '23 at 11:35
Not really - the reasons they have no common factor is more complicated than this, and this approach can't be "fixed," as far as I can tell. – Thomas Andrews Dec 29 '23 at 19:01
In particular, you use a rule which is just blatantly wrong - that an even and an odd number cannot have a common factor. There is no way to replace that step with anything useful. – Thomas Andrews Dec 29 '23 at 19:04

Distinct Fermat numbers are coprime without recursion

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