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This question often comes in my mind when doing exercices in elementary number theory:

Is every non-square integer a primitive root modulo some odd prime?

This would make many exercices much easier. Unfortunately I seem unable to discover anything interesting which may lead to an answer.

It seems likely to me that this is true. If $n\equiv2\pmod3$ then it's a primitive root modulo $3$. If $n\equiv2,3\pmod5$, it's a primitive root modulo $5$. If we would continue like this, my guess is that any non-square $n$ will satify at least one of these congruences.

This being difficult, I began considering a simplified question:

Is every non-square integer a quadratic non-residue modulo some prime?

Or equivalently,

If an integer is a square modulo every prime, then is it a square itself?

The second form seems easier to approach, however I still can't find anything helpful.

Bart Michels
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    The first question is related to http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots – gammatester Jan 21 '14 at 11:30
  • Interesting. However that conjecture asks for infinitely many primes, which is less likely and not really what I'm looking for. – Bart Michels Jan 21 '14 at 11:38
  • Related: https://math.stackexchange.com/questions/6976, https://mathoverflow.net/questions/135820, https://math.stackexchange.com/questions/1250456 – Watson Nov 23 '18 at 20:00

2 Answers2

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Let $n$ be a non-square. Write $n=a^2b$ with $b\ne 1$ square-free. Write $b=p_1\cdot\ldots\cdot p_m$ as product of disctinct primes with $m\ge 1$. For primes $q> n$ the factor $a^2$ can be ignored, so we have that $\left( \frac bq\right)=1$ for almost all primes $q$. According to quadratic reciprocity law, $ \left(\frac{b}{q}\right)$ is determined by $q\bmod 8b$. Also, there is at least one residue $d\bmod 8b$ for which $q\equiv d\pmod{8b}$ implies $ \left(\frac{b}{q}\right)=-1$ (e.g. ensure $\left(\frac d{p_1}\right)=-1$ and $\left(\frac d{p_i}\right)=+1$ for all other $i$ and use the chinese remainder theorem). Especially, $d$ is relatively prime to $8b$ so that by Dirichlet there exist infinitely many primes $q$ with $q\equiv d\pmod{8b}$. For such a $q$ with $q>n$ we conclude that $n$ is not a square modulo $q$.

Hagen von Eitzen
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    Oh yes, I didn't expect it to be that easy. With "According to quadratic reciprocity..." I suppose you mean applying QR to all primes $p_1,\ldots,p_m$ where $q$ is determined modulo $p_k$ and by the parity of $\frac{(q-1)(p_k-1)}4$ and thus certainly modulo $8p_k$. I like this way of roughly handling with primes using CRT and Dirichlet ;). I believe my main question will be much more diffuclt, however. – Bart Michels Jan 21 '14 at 12:19
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Your "equivalent" question seems trivial and therefore not equivalent.

"If an integer is a square modulo every prime, then is it a square itself?"

Given an integer n, there is a prime p > n. If n is a square modulo every prime, including p, then n = n (modulo p) is a square, so n is a square.

BTW. I wrote a short program that checked whether every non-square integer n <= N is the primitive root modulo some prime p > N, with N = 40 x 10^9. It found that actually every non-square n <= N is a primitive root modulo some prime p with N < p <= N + 1357, which is just the 55th prime > N. This seems to indicate that it is more promising to look for a proof than for a counterexample.

gnasher729
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    There is a mistake in your reasoning. $n<p$ being a quadratic residue modulo $p$ does not imply that $n$ is a square. Take for example $n=2$, $p=7$: $2$ is a square modulo $7$ because $2\equiv3^2\pmod7$, but $2$ isn't a square itself. – Bart Michels May 11 '14 at 09:01