Suppose $p \divides n$. Then for all $m \in \Z$: {{begin-eqn}} {{eqn | l = m \perp n       | o = \implies       | r = m \perp n \land m \perp p       | c = Law of Identity; Divisor of One of Coprime Numbers is Coprime to Other }} {{eqn | o = \implies       | r = m \perp p^k n        | c = Integer Coprime to all Factors is Coprime to Whole }} {{eqn | o = \implies       | r = m \perp n       | c = Divisor of One of Coprime Numbers is Coprime to Other }} {{eqn | ll = \leadsto       | l = m \perp p^k n       | o = \iff       | r = m \perp n       | c = {{Defof|Biconditional}} }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map {\Phi_{p^k n} } x       | r = \prod_{\zeta} \paren {x - \zeta}       | c = where the product runs over all primitive complex $p^k n$th roots of unity }} {{eqn | r = \prod_{\substack {1 \mathop \le m \mathop \le p^k n \\ \gcd \set {m, p^k n} = 1} } \paren {x - \map \exp {\frac {2 \pi i m} {p^k n} } }       | c = Condition for Complex Root of Unity to be Primitive }} {{eqn | r = \prod_{\substack {1 \mathop \le m \mathop \le p^k n \\ \gcd \set {m, n} = 1} } \paren {x - \map \exp {\frac {2 \pi i m} {p^k n} } }       | c = as $m \perp p^k n \iff m \perp n$ }} {{eqn | r = \prod_{q \mathop = 0}^{p^k - 1} \prod_{\substack {1 \mathop \le r \mathop \le n \\ \gcd \set {q n + r, n} = 1} } \paren {x - \map \exp {\frac {2 \pi i \paren {q n + r} } {p^k n} } }       | c = Writing $m = q n + r$ by Division Theorem }} {{eqn | r = \prod_{\substack {1 \mathop \le r \mathop \le n \\ \gcd \set {r, n} = 1} } \prod_{q \mathop = 0}^{p^k - 1} \paren {x - \map \exp {\frac {2 \pi i} {p^k} }^q \map \exp {\frac {2 \pi i r} {p^k n} } }       | c = rearranging; GCD with Remainder }} {{eqn | r = \prod_{\substack {1 \mathop \le r \mathop \le n \\ \gcd \set {r, n} = 1} } \paren {x^{p^k} - \map \exp {\frac {2 \pi i r} n} }       | c = Factorisation of $z^n - a$ }} {{eqn | r = \prod_{\zeta} \paren {x^{p^k} - \zeta}       | c = where the product runs over all primitive complex $n$th roots of unity; Condition for Complex Root of Unity to be Primitive }} {{eqn | r = \map {\Phi_n} {x^{p^k} }       | c = {{Defof|Cyclotomic Polynomial}} }} {{end-eqn}} {{qed|lemma}} Now suppose $p \nmid n$. We still have $p \divides p n$. Write $p^k = p^{k - 1} p n$. Notice that the result we proved above holds trivially for $k = 0$: :$\map {\Phi_{p^0 n} } x = \map {\Phi_n } x = \map {\Phi_n } {x^1} = \map {\Phi_n } {x^{p^0}}$ Hence from the above: :$\map {\Phi_{p^k n} } x = \map {\Phi_{p n}} {x^{p^{k - 1}}}$ We need the following result: :the sets $\set {m \in \Z: m \perp p n}$ and $\set {p r: r \perp n}$ are disjoint and has union $\set {m \in \Z: m \perp n}$ First to show that they are indeed disjoint: Suppose $x \in \set {p r: r \perp n}$. Then $p \divides x$. Since $p \divides p n$: :$x \not \perp p n$ and thus: :$x \notin \set {m \in \Z: m \perp p n}$ Hence the sets are disjoint. Now we show that their union is indeed $\set {m \in \Z: m \perp n}$. By Divisor of One of Coprime Numbers is Coprime to Other: :$\forall m \in \Z: m \perp p n \implies \paren {m \perp p \land m \perp n}$ This gives: :$\set {m \in \Z: m \perp p n} \subseteq \set {m \in \Z: m \perp n}$ Let $x \in \set {p r: r \perp n}$. We are given that $p \perp n$. By Integer Coprime to all Factors is Coprime to Whole: :$x \perp n$ Hence $x \in \set {m \in \Z: m \perp n}$. This gives: :$\set {p r: r \perp n} \subseteq \set {m \in \Z: m \perp n}$ By Union of Subsets is Subset: :$\set {m \in \Z: m \perp p n} \cup \set {p r: r \perp n} \subseteq \set {m \in \Z: m \perp n}$ For the other direction, we let $x \notin \set {m \in \Z: m \perp p n} \cup \set {p r: r \perp n}$. Then by De Morgan's Laws (Set Theory)/Set Complement: :$x \in \set {m \in \Z: m \not \perp p n} \cap \set {p r: r \not \perp n}$. By definition of intersection: :$x \in \set {p r: r \not \perp n}$ Thus: :$\exists d \in \Z: d > 1: d \divides r \divides x \land d \divides n$ Therefore $x \not \perp n$. This gives: :$x \notin \set {m \in \Z: m \perp n}$ Hence: :$\set {m \in \Z: m \perp n} \subseteq \set {m \in \Z: m \perp p n} \cup \set {p r: r \perp n}$ and we have our result by definition of set equality. Therefore: {{begin-eqn}} {{eqn | l = \map {\Phi_{p n} } {x^{p^{k - 1} } }       | r = \prod_{\zeta} \paren {x^{p^{k - 1} } - \zeta}       | c = where the product runs over all primitive complex $p n$th roots of unity }} {{eqn | r = \prod_{\substack {1 \mathop \le m \mathop \le p n \\ m \mathop \perp p n} } \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i m} {p n} } }       | c = Condition for Complex Root of Unity to be Primitive }} {{eqn | r = \prod_{\substack {1 \mathop \le m \mathop \le p n \\ m \mathop \perp n} } \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i m} {p n} } } / \prod_{\substack {1 \mathop \le p r \mathop \le p n \\ \gcd \set {r, n} = 1} } \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i m} {p n} } }       | c = from above }} {{eqn | r = \prod_{\substack {1 \mathop \le m \mathop \le p n \\ m \mathop \perp n} } \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i m} {p n} } } / \prod_{\zeta} \paren {x^{p^k} - \zeta}       | c = where the product runs over all primitive complex $n$th roots of unity; Condition for Complex Root of Unity to be Primitive }} {{eqn | r = \prod_{\substack {1 \mathop \le m \mathop \le p n \\ m \mathop \perp n} } \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i m} {p n} } } / \map {\Phi_n} {x^{p^{k - 1} } }       | c = {{Defof|Cyclotomic Polynomial}} }} {{eqn | r = \prod_{q \mathop = 0}^{p - 1} \prod_{\substack {1 \mathop \le r \mathop \le n \\ \gcd \set {q n + r, n} = 1} } \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i \paren {q n + r} } {p n} } } / \map {\Phi_n} {x^{p^{k - 1} } }       | c = Writing $m = q n + r$ by Division Theorem }} {{eqn | r = \prod_{\substack {1 \mathop \le r \mathop \le n \\ \gcd \set {r, n} = 1} } \prod_{q \mathop = 0}^{p - 1} \paren {x^{p^{k - 1} } - \map \exp {\frac {2 \pi i} p}^q \map \exp {\frac {2 \pi i r} {p n} } } / \map {\Phi_n} {x^{p^{k - 1} } }       | c = rearranging; GCD with Remainder }} {{eqn | r = \prod_{\substack {1 \mathop \le r \mathop \le n \\ \gcd \set {r, n} = 1} } \paren {x^{p^k} - \map \exp {\frac {2 \pi i r} n} } / \map {\Phi_n} {x^{p^{k - 1} } }       | c = Factorisation of $z^n - a$ }} {{eqn | r = \prod_{\zeta} \paren {x^{p^k} - \zeta} / \map {\Phi_n} {x^{p^{k - 1} } }       | c = where the product runs over all primitive complex $n$th roots of unity; Condition for Complex Root of Unity to be Primitive }} {{eqn | r = \map {\Phi_n} {x^{p^k} } / \map {\Phi_n} {x^{p^{k - 1} } }       | c = {{Defof|Cyclotomic Polynomial}} }} {{end-eqn}} as required. {{qed}} Category:Cyclotomic Polynomials bfpzg3nvi0k44pocm217165uhcntqfo