Zsigmondy's Theorem for Sums
Theorem
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
with the following exception:
- $n = 3$, $a = 2$, $b = 1$
Outline of Proof
We apply Zsigmondy's Theorem to $a^{2 n} - b^{2 n}$.
Proof
By Zsigmondy's Theorem, there exists a prime divisor $p$ of $a^{2 n} - b^{2 n}$ which does not divide $a^k - b^k$ for all $k < 2 n$ unless:
- $n = 1$ and $a + b$ is a power of $2$
- $n = 3$, $a = 2$, $b = 1$
In particular, $p$ does not divide $a^{2 k} - b^{2 k} = \paren {a^k - b^k} \paren {a^k + b^k}$ for $k < n$.
It remains to check the case $n = 1$ and $a + b$ a power of $2$.
We have to show that $a^2 + b^2$ has an odd prime divisor.
Since $a$ and $b$ are coprime, both $a$ and $b$ are odd.
By Square Modulo 4, $a^2 + b^2 \equiv 2 \pmod 4$.
Because $a > b > 0$, $a^2 + b^2 > 2$.
But $4 \divides 2^k$ for $k > 1$.
Thus $a^2 + b^2$ is not a power of $2$.
Hence $a^2 + b^2$ has an odd prime divisor.
$\blacksquare$
Source of Name
This entry was named for Karl Zsigmondy.