$x+1$ is a factor of $x^n+x^{n-1}+...+x+1$

Question

Prove the following are true in $A[x]$ for any ring A: For any odd $n$,

$x+1$ is a factor of $x^n+1$

$x+1$ is a factor of $x^n+x^{n-1}+...+x+1$.

So I believe we do long division and get that $x^n+1=(x+1)(x^{n-1}+...+x+1)$.I don't know that my math is correct here but then this would show $x+1$ is a factor. Then for 2, $x^n+x^{n-1}+...+x+1= (x^n)+(x^{n-1}+...+x+1)$. I think I would do long division again but I'm not sure with what pieces.

Am I on the right track?

https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd — lab bhattacharjee, Nov 06 '17 at 14:29
Check your long division for $n=3$. You need some minus signs in the second factor. — Ethan Bolker, Nov 06 '17 at 14:30

score 2 · Accepted Answer · answered Nov 06 '17 at 14:31

2

$x+1$ is a factor of $P(x)$ if $P(-1)=0$

As $(-1)^n+1=0$ for odd $n$ $x+1$ divides $x^n+1$

in a similar we can see that $x^n+\ldots+1$ is divisible by $x+1$ for odd $n$

Hope this is useful

answered Nov 06 '17 at 14:31

Raffaele

26,371

score 0 · Answer 2 · answered Nov 06 '17 at 14:30

i believe here is more illustrative to appñy the notion of congruences, or in this case, the factor theorem. Remember, if a polynomial divides another, its roots are also roots of the second. substitute -1 and you will see. also, this helps you build the quotient.

$x+1$ is a factor of $x^n+x^{n-1}+...+x+1$

2 Answers2