Challenge Question

Prove that for ,

Note: There is a very simple solution (in terms of not requiring the use of theorems, not neccesarily in terms of length )

Moderator approved. CB

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- October 3rd 2010, 06:52 AMDefunktFermat Numbers
**Challenge Question**

Prove that for ,

Note: There is a very simple solution (in terms of not requiring the use of theorems, not neccesarily in terms of length )

Moderator approved. CB - October 3rd 2010, 07:22 AMundefined
It has been discussed on the forum before.

http://www.mathhelpforum.com/math-he...ms-144720.html

Note: LaTeX colour tags don't work now, but they used to, hence the words "red" and "black" in my LaTeX.

Note #2: Since that thread is kind of long, you can find the main part of the proof here

http://en.wikipedia.org/wiki/Fermat_...sic_properties - October 3rd 2010, 08:32 AMDefunkt
Oh.. (Sadsmile)

Well, here's my proof, which is a little bit different:

__Spoiler__: - October 3rd 2010, 05:59 PMmelese
Denote by the -th Fermat Number. Observe that

**(1)**

.

What happens? At each stage we take the rightmost term and factor it into: ; continuing this way gives the above factorization of .

Now we can write**(1)**in the following form:**(2)**.

Suppose that . Then by**(2)**, , where is a positive integer. It follows that any common divisor of and must divide , and therefore is equal to or . Fermat Numbers are odd, so the only common divisor they have is .

Incidentally, the result of this problem is one of the proofs that there are infinitely many prime numbers.... - January 4th 2011, 04:29 AMPaulRS
Let's assume , then there's a prime that divides both and .

Thus and but without loss of generality -by symmetry- we may assume that and then which implies thus which is abusrd since both numbers are odd.

**Remark**: In fact one may observe that implies that the order of 2 module p is so applying this to both m and n yields