Let $F_n=2^{2^n}+1$ be a Fermat prime and let $a\in\Bbb{Z}$ such that $F_n\not| a$ and $a$ is not a quadratic residue modulo $F_n$.
I want to show that the class $a+(F_n)$ generates the group of units $(\Bbb{Z}/F_n\Bbb{Z})^\times$.
By the assumptions on $a$, the Legendre symbol $\left(\frac{a}{F_n}\right)=-1$ by definition. Now, there is a proposition by Euler that says that if $p>2$ is a prime and $a\in\Bbb{Z}$, then $a^{\frac{p-1}{2}}\equiv\left(\frac{a}{p}\right)\mod p$. Applied to this case I get that $a^{\frac{F_n-1}{2}}\equiv -1\mod F_n$, i.e. $a^{2^{2^n-1}}\equiv -1(\mod 2^{2^n}+1)$.
I've already used the assumptions on $a$, so I guess I should conclude using that $F_n=2^{2^n}+1$, but I can't find the way.
The group is cyclic of order $2^m$ where $m=2^n$. It has $2^{m-1}$ generators, those elements which are not squares (an element of a cyclic group of order $r$ is a generator iff it is not a $p$-th power for any prime factor $p$ of $m$). So the generators are the quadratic non-residues.