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Given a positive integer n, how would one show that there are infinitely many primes p of the form 4k + 3 that have Legendre symbol (n/p) = -1? From the comments I have received thus far, it has been suggested to use either the Cebotorev Density Theorem or Quadratic Reciprocity to show there infinitely such primes with (n/p) = (-1/p) = -1, but I don't see how and welcome any suggestions about this. I am asking the question for any positive integer n, though I am primarily interested in the case where n is the product of two prime numbers, where it does not happen that both of these prime numbers are congruent to 3 mod 4.

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  • $\begingroup$ As per Bruno Joyal's answer, this question is false as stated. However it is not hard to show that $(n/p) = (-1/p) = -1$ for infinitely many primes $p$ so long as $n$ is not a square. Furthermore, the density of such primes is $1/4$ unless $n$ is of the form $-k^2$, in which case the density is $1/2$. $\endgroup$
    – Erick Wong
    Commented Jan 31, 2014 at 5:29
  • $\begingroup$ Thank you Erick--this is along the lines of what I wanted to know, but can you refer me to where I can find a proof of your statement that (n/p) = (-1/p) = -1 for infinitely many primes p as long as n is not a square, or perhaps briefly outline a proof for me? Thanks. $\endgroup$ Commented Jan 31, 2014 at 5:41
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    $\begingroup$ @ElliotBenjamin Use the Cebotarev density theorem, or quadratic reciprocity. $\endgroup$ Commented Jan 31, 2014 at 6:02
  • $\begingroup$ Thanks Bruno--I'll try that. $\endgroup$ Commented Jan 31, 2014 at 6:12
  • $\begingroup$ @ElliotBenjamin My pleasure! By the way, you should ping users using @ if you want them to be notified of your comment replies. $\endgroup$ Commented Jan 31, 2014 at 6:19

2 Answers 2

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If $n$ is a square, then $(n/p)=1$ always.

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    $\begingroup$ +1. Always except for the finitely many primes dividing $n$. $\endgroup$
    – Erick Wong
    Commented Jan 31, 2014 at 5:25
  • $\begingroup$ @ErickWong Indeed, good eye. I left it out because it had less of a punch ;) $\endgroup$ Commented Jan 31, 2014 at 5:25
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If $n=-1$ then $(n/p)= -1$ for all $4k+3$ primes

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    $\begingroup$ This isn't really an answer, it's a special case. $\endgroup$
    – Erick Wong
    Commented Jan 31, 2014 at 5:26
  • $\begingroup$ I understand. The OP's question is rather vague. Does he want the condition on $n$, or a specific example? The statement is not true in general as Bruno Joyal showed. $\endgroup$
    – user44197
    Commented Jan 31, 2014 at 5:27
  • $\begingroup$ I should probably be more specific for what I want to know. Given n = the product of two primes, such that it's not the case that both primes are congruent to 3 mod 4, and p is a prime congruent to 3 mod 4, why are there infinitely many primes p such that the Legendre symbol (n/p) = -1? $\endgroup$ Commented Jan 31, 2014 at 5:36

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