Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In order to show $$\sum_{0\leq k\leq p-1}\left(\frac{ak^2+bk+c}{p}\right)=\left(\frac{a}{p}\right)$$ with $p$ a prime number greater and equal to 3 and $gcd(b^2-4ac,p)=1$, I started to set $d=b^2-4ac$, and $x=2ak+b$, then the equation becomes $$\sum_{0\leq x\leq p-1}\left(\frac{x^2-d}{p}\right)=\left(\frac{4a^2}{p}\right).$$

Now use the property of Legendre symbol: $$\sum_{0\leq x\leq p-1}\left(\frac{x^2-d}{p}\right)=\sum_{0\leq y\leq p-1}(1+\left(\frac{y+d}{p}\right))\left(\frac{y}{p}\right)=\sum_{0\leq y\leq p-1}\left(\frac{y+d}{p}\right)\left(\frac{y}{p}\right)=-1.$$

However, for the RHS $\left(\frac{4a^2}{p}\right)$ doesn't seems to be equal -1, or is it? Did I make any mistakes in my proof?

share|cite|improve this question
Am I missing something obvious? Why did you change the right hand side of the equation after your substitution? – Arturo Magidin Nov 1 '11 at 4:37
because I times both sides $\left(\frac{4a}{p}\right)$ in order to transform $(ak^2+bk+c)$ into $x^2-d$ – Rob Nov 1 '11 at 4:42
up vote 2 down vote accepted

The thing you are trying to show, with $a=1$, $b=0$, $c=2$, $p=3$, gives $-1$ on the left but 1 on the right, so you are trying to prove a falsehood.

share|cite|improve this answer
Thanks for pointing that out! – Rob Nov 1 '11 at 4:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.