1. ## cycloatomic poly.

See attachment. Thanks.

2. Originally Posted by bigb

Suppose that $\displaystyle n$ is a positive integer, and $\displaystyle m$ is the largest odd divisor of $\displaystyle n.$ Show that $\displaystyle x^n + 1$ factors as a product of $\displaystyle m$ irreducible polynomials.

Hint: $\displaystyle x^n + 1 = \frac{x^{2n}-1}{x^n - 1}$.
this is false! the correct statement is: $\displaystyle x^n+1$ factors as a product of $\displaystyle \tau(m)$ irreducible polynomials, where $\displaystyle \tau(m)$ is the number of divisors of $\displaystyle m$. here is why: so we have $\displaystyle n=2^km,$ for some

integer $\displaystyle k \geq 0.$ see that $\displaystyle A=\{2^{k+1}d: \ d \mid m \}$ is exactly the set of those divisors of $\displaystyle 2n$ which do not divide $\displaystyle n.$ obviously: $\displaystyle |A|=\tau(m).$ now let $\displaystyle \Phi_d(x)$ be the d-th cyclotomic polynomial. then:

$\displaystyle x^n + 1 = \frac{x^{2n} - 1}{x^n - 1} = \frac{\prod_{d \mid 2n} \Phi_d(x)}{\prod_{d \mid n}\Phi_d(x)}=\prod_{r \in A} \Phi_r(x). \ \ \Box$