Is it possible to prove that $3$ is a primitive root of any Fermat prime without quadratic reciprocity? Browsing around online, I find a handful of proofs that $3$ is always a primitive root of any Fermat prime $2^n+1$. One particular proof is found in problems 4 and 5 here. Another proof I found on a page at SUNYSB makes use of Pepin's test whose proof seems to require quadratic reciprocity. 
So out of curiosity, is it possible to prove that any Fermat prime has $3$ as a primitive root without use of quadratic reciprocity? 
The few observations I do know is that if $2^n+1$ is prime for $n\gt 1$, then $n$ is actually a power of $2$, and that in this case $U(\mathbb{Z}/p\mathbb{Z})$ has order $2^n$, so the order of $3$ must also be a power of $2$, but I couldn't get much further than that. 
 A: Hint from Ireland/Rosen which avoids explicitly using quadratic reciprocity:

If $3$ is not a primitive element, show that $3$ is congruent to a square.  Use exercise $4$ (which states that if $q$ is a prime and $q \equiv 1 \pmod{4}$, then $x$ is a quadratic residue $\bmod q$ if and only if $-x$ is a quadratic residue $\bmod{q}$) to show that there is a number $a$ such that $-3 \equiv a^2 \pmod{p}$.  Now solve $2u \equiv - 1 + a \pmod{p}$ and show that $u$ has order $3$.  This would imply that $p \equiv 1 \pmod{3}$, which cannot be true.

A: Let $n$ denote the number $2^{2^{n'}}+1$, $g$ be a primitive root modulo $n$, and let $n(a)$ denote the order of $a$ modulo $n$, for an integer $a$.  Since $a^{n(a)}\equiv g^{n-1} \pmod{n}$, if $n$ is a prime, $a\equiv g^{(n-1)/n(a)} \pmod{n}$; hence $a$ is a quadratic residue modulo $n$ if, and only if, (n-1)/n(a) is even.  Take for $n$ the number $2^{2^{n'}}+1$, we see that $a$ is a quadratic residue modulo $n$ if and only if $a$ is not a primitive root modulo $n$.
As $2^{2^{n'}}+1\equiv 2 \pmod{3}$, for every $n'$>1, $n$ is of the form $3x+2$.  If 3 is a quadratic residue modulo $n$, however, $n$ must divide a number of the form $x^2-3y^2$, hence $\equiv 1 \pmod{3}$, which is obviously a contradiction.  Therefore 3 is not a quadratic residue modulo $n$, thus 3 is a primitive root modulo $n$.
Please forgive me for some typing error.
Thanks in any case for paying attention.  
