The question is in the title, and I do not really have anything to add. Nevertheless I had to write something here in order to be able to ask the question. Thanks.
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Of course. Take a quadratic nonresidue $1\leq n\leq p-1$, then some prime divisor $\ell$ of $n$ will be a quadratic nonresidue. See this MO question for what is known about number fields. |
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Erdos conjectured that for any sufficiently large prime $p$ there is a primitive root |
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I think the answer is obvious. Since $$\sum_{1\leq n\leq p-1}\left(\frac{n}{p}\right)=0$$, there must exist a positive integer $n\leq p-1$, such that $(\frac{n}{p})=-1$, or else the summation above must be equal to $p-1$. Of course, maybe $n$ is not a prime, however there always be a prime factor $\ell$ of $n$ such that $(\frac{\ell}{p})=-1$. |
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Slightly different in emphasis, the smallest quadratic nonresidue is in fact prime, as the product of residues is another residue. |
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It is actually quite easy to prove that if $p>3$, then there are at least $2$ primes less than $p$ which are quadratic non-residues. Indeed, assume there were only one, say $q$. Then every $n$ between $1$ and $p-1$ which is not multiple of $q$ is a quadratic residue. Since you have at most $(p-1)/q$ multiples of $q$, and exactly $(p-1)/2$ quadratic residues, this implies $q=2$ and moreover $p=3$ (since otherwise you would get too many quadratic residues: every odd number between $1$ and $p-1$, together with $4$). |
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