How can I show that $(x-1)(x^2-1)$ divides the polynomial $(x^n-1)(x^{n+1}-1)$?

Question

How can I show that $(x-1)(x^2-1)$ divides the polynomial $(x^n-1)(x^{n+1}-1)$?

Answer 1

• Show/recall that $x-1$ divides $x^k - 1$.
• Use that to show $x^2 - 1 $ divides $(x^2)^k -1$, that is $x^2 - 1$ divides $x^m -1$ for even $m$.
• Distinguish cases according to $n$ odd and even.

Answer 2

$$x-1|x^n-1$$ Now, if $n$ is odd $x+1|x^{n+1}-1$, if $n$ is even then $x-1|x^n-1$. Thus the result follows.

Answer 3

The roots of $f(x):=(x-1)(x^2-1)$ are the multiset $\{1,1,-1\}$. One of $n$ or $n+1$ is even. Assume without loss of generality $n$ is even so that $1$ and $-1$ are roots of $x^n-1$. The other root $1$ is then a root of $x^{n+1}-1$.

So all the roots of $f(x)$ are roots of $g(x):=(x^n-1)(x^{n+1}-1)$ and so by the Factor Theorem all the factors of $f(x)$ are factors of $g(x)$; i.e. $f(x)$ divides $g(x)$.

Answer 4

only trouble is showing that $(x+1)$ divides $f(x) = (x^n - 1)(x^{n+1} -1).$ but $$f(-1) = ((-1)^{n} - 1)((-1)^{n+1} - 1)$$ one of $n$ or $n+1$ must be even therefore $f(-1) = 0$ and $f$ is divisible by $(x+1).$