We call a prime number '''primitive''' if it divides $a^n - b^n$ but not $a^k - b^k$ for any $k < n$. Let $\map {\Phi_n} {x, y}$ denote the $n$th homogeneous cyclotomic polynomial. By Product of Cyclotomic Polynomials: :$a^n - b^n = \displaystyle \prod_{d \mathop \divides n} \map {\Phi_d} {a, b}$ Thus any primitive prime divisor is a divisor of $\map {\Phi_n} {a, b}$. We start by investigating to which extent the converse is true. === Upper bound on nonprimitive prime divisors === Let $p$ be a prime divisor of $\map {\Phi_n} {a, b}$ which is not primitive. Then there exists $k \mid n$ with $k < n$ such that $p \divides a^k - b^k$. From Product of Cyclotomic Polynomials: :$p \divides \dfrac {a^n - b^n} {a^k - b^k}$ By P-adic Valuation of Difference of Powers with Coprime Exponent: :$p \divides \dfrac n k$ In particular: :$p \divides n$ Let $n = p^\alpha q$ with $p \nmid q$. By Cyclotomic Polynomial of Index times Prime Power: :$p \divides \map {\Phi_n} {a, b} \divides \map {\Phi_q} {a^{p^\alpha}, b^{p^\alpha} }$ By Fermat's Little Theorem: :$p \divides \map {\Phi_q} {a, b}$ By Prime Divisors of Cyclotomic Polynomials: :$p \equiv 1 \pmod q$ Thus $p$ is the largest prime divisor of $n$. Next, we determine the valuation of $p$ in $\map {\Phi_n} {a, b}$. By Multiplicative Order of Roots of Cyclotomic Polynomial Modulo Prime, if $d \divides n$ with $p \divides a^d - b^d$, then $q \divides d$. By the Möbius Inversion Formula for Cyclotomic Polynomials: :$\map {\Phi_n} {a, b} = \displaystyle \prod_{d \mathop \divides n} \paren {a^d - b^d}^{\map \mu {n/d} }$ Taking the valuation: :$\map {\nu_p} {\map {\Phi_n} {a, b} } = \displaystyle \sum_{q \mathop \divides d \mathop \divides n} \map \mu {n / d} \cdot \map {\nu_p} {a^d - b^d}$ The only nonzero terms are for $d = n$ and $d = n / p$. Thus: :$\map {\nu_p} {\map {\Phi_n} {a, b} } = \map {\nu_p} {\dfrac {a^n - b^n} {a^{n / p} - b^{n / p} } }$ By Fermat's Little Theorem: :$p \divides a^{n/p} - b^{n/p}$ If $p > 2$, then by the Lifting The Exponent Lemma: :$\map {\nu_p} {\map {\Phi_n} {a, b} } = 1$ If $p = 2$, then $n$ is a power of $2$. By Cyclotomic Polynomial of Index Power of Two: :$\map {\Phi_n} {a, b} = a^{n/2} + b^{n/2}$ If $n > 2$, then by Square Modulo 4: :$\map {\Phi_n} {a, b} \equiv 2 \pmod 4$ Thus: :$\map {\nu_p} {\map {\Phi_n} {a, b} } = 1$ If $n = 2$ and $a + b$ is not a power of $2$, then any odd prime divisor of $a + b$ is primitive, and the theorem is true. We may thus suppose $\map {\nu_p} {\map {\Phi_n} {a, b} } = 1$. === Lower bound on primitive prime divisors === If $\map {\Phi_n} {a, b}$ has no nonprimitive prime divisors, it suffices to show that $\size {\map {\Phi_n} {a, b} } > 1$. By Trivial Estimate for Cyclotomic Polynomials: :$\size {\map {\Phi_n} {a, b} } \ge \paren {a - b}^{\map \phi n} \ge 1$ with equality {{iff}} $n = 1$ and $a - b = 1$. If $\map {\Phi_n} {a, b}$ has a nonprimitive prime divisor $p \divides n$, let $n = p^\alpha q$ with $p \nmid q$. We have to show that $\map {\Phi_n} {a, b} > p$. If $\alpha > 1$, then: {{begin-eqn}} {{eqn | l = \map {\Phi_n} {a, b}       | r = \map {\Phi_{p^{\alpha - 1} q} } {a^p, b^p}       | c = Cyclotomic Polynomial of Index times Prime Power }} {{eqn | o = \ge       | r = a^p - b^p       | c = Trivial Estimate for Cyclotomic Polynomials }} {{eqn | o = \ge       | r = \paren {b + 1}^p - b^p }} {{eqn | o = >       | r = b p       | c = Binomial Theorem }} {{eqn | o = \ge       | r = p }} {{end-eqn}} and the proof is complete. Let $\alpha = 1$. Then $n = p q$. If $a - b > 1$ then: {{begin-eqn}} {{eqn | l = \map {\Phi_n} {a, b}       | o = \ge       | r = \paren {a - b}^{\map \phi n}       | c = Trivial Estimate for Cyclotomic Polynomials }} {{eqn | o = \ge       | r = 2^{p - 1}       | c = $a - b \ge 2$ }} {{eqn | o = \ge       | r = p }} {{end-eqn}} with equality only if $n = 1$, in which case the theorem is trivially true. If $a - b = 1$ then: {{begin-eqn}} {{eqn | l = \map {\Phi_n} {a, b}       | r = \frac {\map {\Phi_q} {a^p, b^p} } {\map {\Phi_q} {a, b} }       | c = Cyclotomic Polynomial of Index times Prime Power }} {{eqn | o = \ge       | r = \frac {\paren {a^p - b^p}^{\map \phi q} } {\paren {a + b}^{\map \phi q} }       | c = Trivial Estimate for Cyclotomic Polynomials }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \frac {a^p - b^p} {a + b}       | r = \frac {\paren {b + 1}^p - b^p} {2 b + 1}       | c = $a - b = 1$ }} {{eqn | o = \ge       | r = \frac {\paren {2^p - 1} b} {2 b + 1}       | c = Binomial Theorem }} {{eqn | o = \ge       | r = \frac {2^p - 1} 3 }} {{eqn | o = \ge       | r = 1 }} {{end-eqn}} so: :$\map {\Phi_n} {a, b} \ge \dfrac {2^p - 1} 3$ If $p \ge 5$, then: :$\dfrac {2^p - 1} 3 > p$ and the proof is complete. Let $p = 2$. Then $n = 2^\alpha = 2$. Because $a - b = 1$, any prime divisor of $a + b$ is then primitive. Let $p = 3$. Then $n = 3 q$. From $p \equiv 1 \pmod q$ we have $n = 3$ or $n = 6$. If $n = 3$, then: :$a^n - b^n = a^2 + a b + b^2$ and any prime divisor of $a^2 + a b + b^2$ is primitive. If $n = 6$, then: :$\map {\Phi_n} {a, b} = a^2 - a b + b^2 = b^2 + b + 1$ Thus $\map {\Phi_6} {a, b} > 3$ unless $a = 2$ and $b = 1$. {{qed}} {{Namedfor|Karl Zsigmondy|cat = Zsigmondy}}