Can an element be a quadratic residue and a generator (mod p)? i.e. is is possible for 


*

*$g$ to be a generator$\mod{p}$, and

*$g \equiv x^2 \mod{p}$ for some $x$


I'm guessing not, as I think $x$ can't be expressed as a power of $g$, contradicting g being a generator?
 A: If $p$ is an odd prime, then no. $\varphi(p)=p-1$ is even and the multiplicative group $\!\!\!\pmod{p}$ has order $\varphi(p)$.  If $g=x^2$, then $g^{(p-1)/2}=x^{p-1}=1\pmod{p}$. But if $g$ is a generator of the multiplicative group $\!\!\!\pmod{p}$, $g^k\not=1\pmod{p}$ for $0<k<p-1$.
In fact, the only $p$ for which $\varphi(p)$ is odd is $p=2$, and $1=1^2$ is a generator for the multiplicative group $\!\!\!\pmod{2}$.  For all other $p$, $g=x^2$ cannot be a generator of the multiplicative group $\!\!\!\pmod{p}$.
A: If you consider $p=2$ to be a prime number (and I wouldn't know how you could defend not doing so) then you will see that $1$ is both a quadratic residue and a generator mod $2$ (admittedly not very spectacular, but true nontheless). This is the only case, as explained the answer by robjohn.
A: A bit late :) but another way to see that this cannot be true for $p>2$ is using the following theorem:

For a prime $p>2$, exactly half of the elements in $\mathbb{Z}^*_p$ are QRs.

If the generator is a QR, then so is every other element.
