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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?

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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
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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.

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Thanks for pointing that out! –  Rob Nov 1 '11 at 4:45
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