Let $n$ be an integer and $p_1$, $p_2$ be primes such that $p_1 \equiv p_2 \pmod{4n}$. Prove $\left(\dfrac{n}{p_1}\right)=\left(\dfrac{n}{p_2}\right)$.

2 cases:

  1. $p_1=4nk_1+1$, $p_2=4nk_2+1$. $\left(\dfrac{n}{p_1}\right) \equiv n^{(p_1-1)/2}\equiv n^{(4nk_1+1-1)/2}\equiv n^{2nk_1} \mod{p_1}$...

  2. $p_1=4nk_1+3$, $p_2=4nk_2+3$.

Not really help...



If $n$ is an odd number. $n=2k+1$.

by the property 2 of the link.

$(\dfrac{p_1}{n})=(\dfrac{p_2}{n}) $, since $p_1 = p_2 + 4nl$.

Then by The law of quadratic reciprocity:

$(\dfrac{n}{p_1})(\dfrac{p_1}{n}) = (-1)^{\frac{n-1}{2}\frac{p_1-1}{2}} = (-1)^{k\frac{p_2+4nl-1}{2}} = (-1)^{k\frac{p_2-1}{2}} = (-1)^{\frac{n-1}{2}\frac{p_2-1}{2}} = (\dfrac{n}{p_2})(\dfrac{p_2}{n})$

So you have


When $n = 2^{\alpha}(2k+1)$, you simply consider $\alpha = 1$.

And for $i =1 ,2$.

$(\dfrac{p_i}{n}) = (\dfrac{p_i}{2^{\alpha}})(\dfrac{p_i}{2k+1}) = (\dfrac{p_i}{2})^{\alpha}(\dfrac{p_i}{2k+1}) $

you may try to see the value of $(\dfrac{p_1}{2})$ and $(\dfrac{p_2}{2})$

  • $\begingroup$ A side: if one of $p_1$, $p_2$ is $2$, then your arguments might fail. So you have to exclude the case that one of $p_1, p_2$ is odd, by dint of the congruence condition. Just for the sake of completeness. Regards. $\endgroup$ – awllower Apr 2 '13 at 6:25

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