Fermat

[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.

[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.

[mathwonk]

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

[robert Ihnot]

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

[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".

[robert Ihnot]

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".

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.

[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.