1. ## cycloatomic poly.

See attachment. Thanks.

2. Originally Posted by bigb

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

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

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

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