Let $p$ be a prime, then is it true that a generator $g$ of $\mathbb{Z}^*_p$ cannot be a quadratic residue modulo $p$? And if yes why?

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    $\begingroup$ Think about it: If $g$ is a quadratic residue then so is any power of $g$. (P.S. $p=2$ is different. You need to consider only odd primes.) $\endgroup$ – Harald Hanche-Olsen Apr 29 '16 at 13:58

Yes, it is true, for odd $p$.

The point is that a generator $g$ has order $p-1$, which is even.
Yet, if $g= a^2$, then $g^{(p-1)/2} = a^{p-1} = 1$, a contradiction to $p-1$ being the order of $g$.

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    $\begingroup$ Very nicely put and amazingly simple. +1 $\endgroup$ – DonAntonio Apr 29 '16 at 20:05

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