Why does a prime p divide a Fermat Number?

[MattcAnderson]

Hi all,
There is an interesting article in Mathematics Magazine (Vol 92, No 4, October 2020).
The title of the article is "Why does a prime p divide a Fermat Number?" They reference fermatsearch.org. They state that a Fermat prime divisor divides one and only one Fermat number.

Regards,
Matt

[LaurV]

Originally Posted by MattcAnderson

Why does a prime p divide a Fermat Number?

Because he can, and he wants to do so.
The "unicity" of factors is known, and trivial to prove.

[R. Gerbicz]

Originally Posted by MattcAnderson

They state that a Fermat prime divisor divides one and only one Fermat number.

I've learnt in this nice way: F(n)=2+F(0)*F(1)*F(2)*...*F(n-1), so if we have a common divisor d of F(n) and F(m), where m<n then d|2, hence d=1 because all F(k) is odd.

[LaurV]

Yep, that's what wikipedia used to say for a while

[R. Gerbicz]

Originally Posted by LaurV

Yep, that's what wikipedia used to say for a while

Really not read, in general I know the shortest proofs. Another way:
If 2^(2^m)==-1 mod d then for n>m taking this to the 2^(n-m)-th power:
2^(2^n)==1 mod d from this F(n)==2 mod d and the rest is the same. This could be even shorter proof when you write down, but needs more knowledge.

[Batalov]

Originally Posted by LaurV

Because he can, and he wants to do so.

Are you sure that this is not the answer for the 'Why did the chicken cross the road?"

Then again, this answer is quite universal, just like "42". I like it.

[LaurV]

That's not a "show off" for the fact I know the math. I may know it, I may not know it. I was barking more in that direction that people come and ask questions whose answers (and more details about subject) could be found by a simple search.

[MattcAnderson]

We know that the Fermat numbers are defined as
F(n) = 2^(2^n) + 1.

Also,
F(0) to F(4) are prime numbers.
F(5) to F(32) are composite numbers.
F(33) to F(35) are of unknown character.
F(36) is composite.

This data is available at -
http://www.prothsearch.com/fermat.html

I spent several years using trial division programs
trying to find the smallest prime factor of F(34).

So far we know that the smallest prime divisor of
F(34) is greater than 7*10^17.

Maybe someone else wants to work on this.
The program (mmff) can be downloaded from
fermatsearch.org

Regards,
Matt

[Batalov]

Originally Posted by MattcAnderson

I spent several years using trial division programs
trying to find the smallest prime factor of F(34).

If you used the programs that most people use (didn't write your own?) - then you don't know which F(n) you will find a factor for.

The stats show that you mostly reserved ranges for N=36.
This means that you can just as well find a factor of F(33) or F(32), if you keep at it.
You are not guaranteed to find the factor of F(34)!