Zsigmondy's Theorem

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Let $a > b > 0$ be coprime positive integers.

Let $n \ge 1$ be a (strictly) positive integer.

Then there is a prime number $p$ such that

$p$ divides $a^n - b^n$
$p$ does not divide $a^k - b^k$ for all $k < n$

with the following exceptions:

$n = 1$ and $a - b = 1$
$n = 2$ and $a + b$ is a power of $2$
$n = 6$, $a = 2$, $b = 1$


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 $d = n / p$.


$\map {\nu_p} {\map {\Phi_n} {a, b} } = \map {\nu_p} {\dfrac {a^n - b^n} {a^{n / p} - b^{n / p} } }$

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:

$\map {\Phi_n} {a, b} \equiv 2 \pmod 4$


$\map {\nu_p} {\map {\Phi_n} {a, b} } = 1$

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 if and only if $n = 1$ 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:

\(\displaystyle \map {\Phi_n} {a, b}\) \(=\) \(\displaystyle \map {\Phi_{p^{\alpha - 1} q} } {a^p, b^p}\) Cyclotomic Polynomial of Index times Prime Power
\(\displaystyle \) \(\ge\) \(\displaystyle a^p - b^p\) Trivial Estimate for Cyclotomic Polynomials
\(\displaystyle \) \(\ge\) \(\displaystyle \paren {b + 1}^p - b^p\)
\(\displaystyle \) \(>\) \(\displaystyle b p\) Binomial Theorem
\(\displaystyle \) \(\ge\) \(\displaystyle p\)

and the proof is complete.

Let $\alpha = 1$.

Then $n = p q$.

If $a - b > 1$ then:

\(\displaystyle \map {\Phi_n} {a, b}\) \(\ge\) \(\displaystyle \paren {a - b}^{\map \phi n}\) Trivial Estimate for Cyclotomic Polynomials
\(\displaystyle \) \(\ge\) \(\displaystyle 2^{p - 1}\) $a - b \ge 2$
\(\displaystyle \) \(\ge\) \(\displaystyle p\)

with equality only if $n = 1$, in which case the theorem is trivially true.

If $a - b = 1$ then:

\(\displaystyle \map {\Phi_n} {a, b}\) \(=\) \(\displaystyle \frac {\map {\Phi_q} {a^p, b^p} } {\map {\Phi_q} {a, b} }\) Cyclotomic Polynomial of Index times Prime Power
\(\displaystyle \) \(\ge\) \(\displaystyle \frac {\paren {a^p - b^p}^{\map \phi q} } {\paren {a + b}^{\map \phi q} }\) Trivial Estimate for Cyclotomic Polynomials

We have:

\(\displaystyle \frac {a^p - b^p} {a + b}\) \(=\) \(\displaystyle \frac {\paren {b + 1}^p - b^p} {2 b + 1}\) $a - b = 1$
\(\displaystyle \) \(\ge\) \(\displaystyle \frac {\paren {2^p - 1} b} {2 b + 1}\) Binomial Theorem
\(\displaystyle \) \(\ge\) \(\displaystyle \frac {2^p - 1} 3\)
\(\displaystyle \) \(\ge\) \(\displaystyle 1\)


$\map {\Phi_n} {a, b} \ge \dfrac {2^p - 1} 3$

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 $b = 1$.


Source of Name

This entry was named for Karl Zsigmondy.

Also see