For any integer $a$ and any two primes $p,q$ with $(p,q)=1$. Prove that if $p \equiv q$ mod $4a$, then $(\frac{a}{p})=(\frac{a}{q})$

I know I need to write $a=(−1)^{e_0} 2^{e_2} p_1 p_2 \cdots p_r$ with $p_i$ odd and possibly equal; and then compute the Legendre symbol and apply the reciprocity laws, I am just unfamiliar with this process and would appreciate any help.


You can show that this is equivalent to the following form of quadratic reciprocity: $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)(q-1)/4}$, the crux of with lies in Gauss's Lemma. To show that this statement implies your's, we just need to show your statement holds when $a$ is prime, due to multiplicativity. I'll leave it to you to show it holds for $a=2$. Otherwise if $a$ is an odd prime then $$\left(\frac{a}{p}\right)=(-1)^{(p-1)(a-1)/4}\left(\frac{a}{q}\right).$$ If $p\equiv q\pmod{4a}$ then $\left(\frac{p}{a}\right)=\left(\frac{q}{a}\right)$, so \begin{align*}\left(\frac{a}{p}\right)=(-1)^{(p-1)(a-1)/4}\left(\frac{q}{a}\right)&=(-1)^{(p-1)(a-1)/4}(-1)^{(q-1)(a-1)/4}\left(\frac{a}{q}\right) \\ &= (-1)^{(a-1)(p+q-2)/4}\left(\frac{a}{q}\right).\end{align*} We have $p+q-2\equiv 0\pmod{4}$, so the result follows. A similar argument works if $p\equiv -q\pmod{4}$, so your original statement can be strengthened to $p\equiv \pm q\pmod{4a}$.

To show the other direction, suppose WLOG that $p>q$ and $p\equiv q\pmod{4}$. We can write $p=q+4a$ for $a\ge 1$, so we get $$\left(\frac{p}{q}\right)=\left(\frac{q+4a}{q}\right)=\left(\frac{a}{q}\right)=\left(\frac{a}{p}\right)=\left(\frac{4a}{p}\right)=\left(\frac{p-q}{p}\right)=\left(\frac{-q}{p}\right)=(-1)^{(p-1)/2}\left(\frac{p}{q}\right).$$

If $p$ is of the shape $4k+1$ then $\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$ which yields the result. Else if $p$ is of the shape $4k+3$, then $q$ is of the shape $4\ell + 3$ also, so $\left(\frac{p}{q}\right)=-\left(\frac{q}{p}\right)$, which also gives the result. If $p$ is of the shape $4k-q$, then $p+q=4a$ so $$\left(\frac{p}{q}\right)=\left(\frac{4a-q}{q}\right)=\left(\frac{a}{q}\right)=\left(\frac{a}{p}\right)=\left(\frac{4a}{p}\right)=\left(\frac{p+q}{p}\right)=\left(\frac{q}{p}\right),$$ which also yields the result since one of $p$ or $q$ must be of the shape $4k+1$. We are done.

  • $\begingroup$ Thank you, I appreciate your help. $\endgroup$ – user99262 Nov 27 '13 at 21:58

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