# Fermat proof

1. Sep 19, 2004

### ascheras

Prove the following is valid only when n is an odd integer.

x^n + 1= (x+1)(x^n - X^n-2 + ... + (x^2 - x + 1).

It's an easy 3 line proof.

2. Sep 19, 2004

### zefram_c

The notation given already implies that n is an integer. From there, it's easy to check that even values of n result in the wrong sign for the leading order term on RHS.

3. Sep 19, 2004

### mathwonk

(x+1) a factor implies x = -1 is a root, but it ain't. (one liner)

4. Sep 19, 2004

### robert Ihnot

The way that is meant is (X^(2n+1)+1) =(X+1)(X^(2n)-X^(2n-1)+X^(2n-2)-+-..+1).

5. Sep 20, 2004

### mathwonk

robert, how can you tell he meant the converse of what he said? what he said was to prove his statement false for even n, not to prove it true for odd n. that at least is how i translate the word "only".

6. Sep 20, 2004

### robert Ihnot

Yes, that is true. Even so statement is not correct for odd n since

(X^3+1) not equal to (X+1)(X^3-X+1) for all X.

Last edited: Sep 20, 2004
7. Sep 21, 2004

### ascheras

Let n= (2^r)*q, where q has no prime divisors. Then you can write (2^n) +1= (y^q) +1, where y=(2^r). Then (2^n) +1= (y+1)(y^(q-1)- y^(q-2) +...+ (y^2) - y+1). Here, y+1= (2^r)+1 >1 and there are two factors. This (2^n)+1 cannot be prime if the other factor is also > 1. That happens unless y^(q-1) +...+ (y^2)- y+1 reduces to 1, i.e. q=1. Therefore (2^n)+1 is prime, which implies that q=1 and n=(2^r) for some r as claimed.