Divison of polynomials

Question

I need to find m and n , ($m,n\in R$ ) so that $P(x)=x^m + x^{m-1} + \dots+x+1$ is divisible by $Q(x)=x^n + x^{n-1} + \dots+x+1$ I have no clue what to do here except that $m>n$ and $n|m$. What should the values of $m$ and $n$ be ?

Answer 1

Hint : if you multiply both polynomials by $X-1$ you get respectively $X^{n+1}-1$ and $X^{m+1}-1$. Now look at the complex roots of these, they are pretty well-known.

Answer 2

If $P(x) = k(x)Q(x)$ then $P(x)(x-1)=k(x)Q(x)(x-1)$ so $$x^{m+1}-1 = k(x)(x^{n+1}-1)$$ and this is well know it is true exactly when $n+1\mid m+1$.

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Anyway you can prove this statement pretty easly.

If $$1+x+...+x^n\mid 1+...+x^m$$ then this is true for all integers so specially for $x=1$ and thus $n+1\mid m+1$.

And vice versa. Write $a=m+1$ and $b=n+1$ then $a=lb$ so $$x^a-1 = (x^b)^l-1=(x^b-1)((x^b)^{l-1}+...+x^b+1)$$ and we are done.

Answer 3

Since $1$ is a root neither of $P$ nor of $Q$, this is the same as asking when $x^{m+1}-1$ divides $x^{n+1}-1$. It's well known that $$ x^k-1=\prod_{d\mid k}\Phi_d(x) $$ where $\Phi_d(x)$ is the $d$-th cyclotomic polynomial, that is, the product $$ \Phi_d(x)=\prod_{\zeta}(x-\zeta) $$ where the product runs over the primitive $d$-th roots of unity. For instance, $$ \Phi_1(x)=x-1,\quad \Phi_2(x)=x+1,\quad \Phi_3(x)=x^2+x+1,\quad \Phi_4(x)=x^2+1 $$ and each $\Phi_d(x)$ is irreducible over $\mathbb{Q}$.

Now put everything together.

The necessary and sufficient condition is $(m+1)\mid(n+1)$