4
$\begingroup$

If $n\in\Bbb Z^+$ is not a square, prove exist infinitely many primes $p$ such that $\left(\frac{n}{p}\right)=-1$.

Note that if $p\nmid n$ and $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, then $\left(\frac{n}{p}\right)=\left(\frac{p_1}{p}\right)^{\alpha_1}\left(\frac{p_2}{p}\right)^{\alpha_2}\cdots \left(\frac{p_k}{p}\right)^{\alpha_k}$

$=\left(\frac{p_1}{p}\right)^{\alpha_1\pmod {2}}\left(\frac{p_2}{p}\right)^{\alpha_2\pmod{2}}\cdots \left(\frac{p_k}{p}\right)^{\alpha_k\pmod{2}}$

So i.e. prove exist infinitely many primes $p$ such that $\left(\frac{q_1}{p}\right)\left(\frac{q_2}{p}\right)\cdots \left(\frac{q_n}{p}\right)=-1$, no matter what primes $q_i$ you take.

$\endgroup$

1 Answer 1

4
$\begingroup$

Given primes $q_1,\ldots,q_n$ pick $\epsilon_,\ldots,\epsilon_n\in\{\pm 1\}$ such that $\prod \epsilon_i=-1$. Then note that $(\frac {q_i}p)=\epsilon_i$ is just a condition for $p\bmod q_i$ (or for $p\bmod 8$ if $q_i=2$). Thus in the end it suffices to have infinitely many primes $p\equiv a\pmod b$ for some $a,b$ with $\gcd(a,b)=1$.

$\endgroup$
1
  • 1
    $\begingroup$ This uses Dirichlet's theorem. We'll see if someone can come up with a simpler solution. $\endgroup$
    – user263326
    Commented Oct 27, 2015 at 7:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .