I was doing a study on Fermat Numbers when I came across this theorem by Édouard Lucas (unproven in my reference material):
Every prime divisor of $F_n = 2^{2^{n}} + 1$ is of the form $k \cdot{2^{n+2}} + 1$ for some $ k \in \mathbb{N}$ and $ n > 1$.
I was able to prove that that it is indeed of the form $K \cdot{2^{n+1}} + 1$ using the following method:
Let $p$ be a prime factor of $F_n = 2^{2^{n}} + 1$. Since $F_n$ is odd, $p$ is odd. $\tag*{}$ $\implies \gcd(2,p) = 1 \\$ $\implies 2^{2^{n}} \equiv -1 \pmod{p} \\$ $\implies (2^{2^{n}})^2 \equiv 1 \pmod{p} \\$ $\implies 2^{2^{n+1}} \equiv 1 \pmod{p} \\$ $\implies \text{ord}_p(2) \mid 2^{n+1} \text{ but } \text{ord}_p(2) \nmid 2^{n} \\$ $\implies \text{ord}_p(2) = 2^{n+1}$ $\tag*{}$ Now using Fermat's Little Theorem: $$2^{p-1} \equiv 1 \pmod{p}$$ $\implies \text{ord}_p(2) \mid p-1 \implies 2^{n+1} \mid p-1 \\$ $\implies p = K \cdot{2^{n+1}} + 1 \text{ where } K \in \mathbb{N}$
But now how do we prove that $K$ is even?
I found a hint here which suggests the use of quadratic residues of $p$. But I have little to no knowledge of quadratic residues and its implications yet.
So it'd be much appreciated if you could walk me through the concept and how it is being used in respect to the problem at hand. A separate proof for the initial problem is also welcome. Thanks.
Remark: I could not find any good material on quadratic residues online, and certainly not on its implications that are being used here.
