Let $F_{n}=2^{2^n}+1$ be a Fermat number. A classic idea using orders and Fermat's Little Theorem shows that a prime divisor $p$ of $F_{n}$ must be of the form $p=k .2^{n+1}+1$. Furthermore, using the fact that such prime has $2$ as a quadratic residue (for $n \geq 2$), we can improve this result to conclude that $p$ has the form $p=k .2^{n+2}+1$ (for a proof, see the theorem of Édouard Lucas here). My question simply concerns when is $k$ an even number and, thus, when can we improve that $n+2$ to a $n+3$.
If $2$ is a quartic residue $\mod p$ ($2 \equiv x^4 \mod p$ for some integer $x$) an analogous argument as the one used in the link above, shows that $p$ has the form $p=s. 2^{n+3}+1$, so $k$ is even, as desired. Now, I'm trying to find out if the reciprocal is valid, so my real question is:
Let $p$ be a divisor of $F_n$. As stated above, $p = k. 2^{n+2}+1$ for some $k$. Suppose that $k$ is an even integer. Then, is it true that $2$ is a quartic residue $\mod p$?
Some computational experiments using this link suggest so. Also, we can use more simple conditions on quartic residues to prove this, like: As $p\equiv 1 \mod 4$, by a very famous result, $p=a^2+b^2$. WLOG, assume $a$ odd and $b$ even. So, $2$ is a quartic residue $\mod p$ if and only if $b \equiv 0 \mod 8$ (this is a theorem by Gauss, referenced here).
Thank you!