Let r,p,q be distinct odd primes. Let 4r divide p-q. Show that

(r/p) = (r/q)

Where (a/b) is the Legendre symbol.

I'm sure we are suppose to use the law of quadratic reciprocity. I don't think this question is suppose to be difficult, but I cannot figure it out!

  • 1
    $\begingroup$ So what do you get when you apply QR to (r/p) for instance? You should at least play around some. Companion to Zev's hint: $$\begin{cases}p\equiv q\bmod r\implies \left(\frac{p}{r}\right)=\left(\frac{q}{r}\right) \\ p\equiv q\bmod 4\implies \frac{p-1}{2}\equiv\frac{q-1}{2}\bmod 2 \end{cases}$$ $\endgroup$
    – anon
    Jun 2, 2013 at 7:35

1 Answer 1


Hint: $$4r\mid (p-q)\iff p\equiv q\bmod r\quad\text{and}\quad p\equiv q\bmod 4$$

  • $\begingroup$ This means (p/r)=(q/r), right? $\endgroup$ Jun 2, 2013 at 7:39
  • $\begingroup$ @John: Just $p\equiv q\bmod r$ alone is enough to know that $(\frac{p}{r})=(\frac{q}{r})$. Do you see how to use the other piece of information? $\endgroup$ Jun 2, 2013 at 7:41
  • $\begingroup$ Yes! Thank you so much. $\endgroup$ Jun 2, 2013 at 7:45
  • $\begingroup$ Glad I could help! $\endgroup$ Jun 2, 2013 at 7:51

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