Divisor of Fermat Number/Refinement by Lucas

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Theorem

Let $F_n$ be a Fermat number.

Let $m$ be divisor of $F_n$.

Let $n \ge 2$.

Then $m$ is in the form:

$k \, 2^{n + 2} + 1$


Proof

It is sufficient to prove the result for prime divisors.

The general argument for all divisors follows from the argument:

$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$

So the product of two factors of the form preserves that form.


Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$.

From Euler's Result:

$\exists q \in \Z: p = q \, 2^{n + 1} + 1$

Since $n \ge 2$, $q \, 2^{n + 1}$ is divisible by $2^{2 + 1} = 8$.

Hence:

$p \equiv 1 \pmod 8$


By Second Supplement to Law of Quadratic Reciprocity:

$\paren {\dfrac 2 p} = 1$

so $2$ is a quadratic residue modulo $p$.

Hence:

$\exists x \in \Z: x^2 = 2 \pmod p$

We have shown $2^{2^n} \equiv -1 \pmod p$ and $2^{2^{n + 1} } \equiv 1 \pmod p$.

By Congruence of Powers:

$x^{2^{n + 1} } \equiv 2^{2^n} \equiv -1 \pmod p$
$x^{2^{n + 2} } \equiv 2^{2^{n + 1}} \equiv 1 \pmod p$


From Integer to Power of Multiple of Order, the order of $x$ modulo $p$ divides $2^{n + 2}$ but not $2^{n + 1}$.

Therefore it must be $2^{n + 2}$.


Hence:

\(\displaystyle \exists k \in \Z: \ \ \) \(\displaystyle \map \phi p\) \(=\) \(\displaystyle k \, 2^{n + 2}\) Corollary to Integer to Power of Multiple of Order
\(\displaystyle p - 1\) \(=\) \(\displaystyle k \, 2^{n + 2}\) Euler Phi Function of Prime
\(\displaystyle p\) \(=\) \(\displaystyle k \, 2^{n + 2} + 1\)

$\blacksquare$


Historical Note

In $1747$, Leonhard Paul Euler proved that a divisor of a Fermat number $F_n$ is always in the form $k \, 2^{n + 1} + 1$.

This was later refined to $k \, 2^{n + 2} + 1$ by François Édouard Anatole Lucas.