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Let $p$ be an odd prime and $p\nmid b^2-4ac$ for integers $a>0,b,c$. Can we show the set $$ \{a\}\cup\{ah^2+bh+c:0\leq h<p\} $$ contains an equal number of quadratic residues and non-residues modulo $p$?

pel
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  • As @MarcBogaerts observed, this is false. But if you replace "set" by "multiset", it should be true. It then boils down to $\left(\dfrac{a}{p}\right) + \sum\limits_{h=0}^{p-1} \left(\dfrac{ah^2+bh+c}{p}\right) = 0$, which I think is a known fact. – darij grinberg Aug 08 '19 at 07:59
  • Ah, there we are: https://math.stackexchange.com/a/1892074/ – darij grinberg Aug 08 '19 at 08:06

2 Answers2

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Contradiction: with $a = 10, b = 5, c = 3, p =97$ the set contains $50$ quadratic residues and $48$ non-residues.

Marc Bogaerts
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  • I get $25$ and $24$ instead, but still a counterexample. – darij grinberg Aug 08 '19 at 07:58
  • I forgot that the values $ah^2+bh+c$ occured twice. – Marc Bogaerts Aug 08 '19 at 08:02
  • ${a}\cup{ah^2+bh+c:0\leq h<p} = {10} \cup {0, 3, 4, 5, 6, 7, 8, 11, 12, 17, 18, 21, 22, 23, 24, 25, 26, 29, 33, 34, 35, 37, 40, 42, 43, 44, 46, 48, 49, 50, 53, 56, 58, 63, 68, 70, 73, 76, 77, 78, 80, 82, 83, 84, 86, 89, 91, 92, 93 }$ This set has 50 elements. Among these, the QRs are the following: ${0, 3, 4, 6, 8, 11, 12, 18, 22, 24, 25, 33, 35, 43, 44, 48, 49, 50, 53, 70, 73, 86, 89, 91, 93}$
    The rest are QNRs. Hence no of QRs = no of QNRs = 25     – PTDS Aug 09 '19 at 05:49
  • @PTDS Indeed, you are right, I miscounted somewhere. – Marc Bogaerts Aug 09 '19 at 08:56
  • @PTDS QRs don't include $0$ by the usual definition. If you do include $0$, then $p = 5$, $a = 1$, $b = 0$ and $c = 1$ is a counterexample. – darij grinberg Aug 09 '19 at 11:45
  • In this case, ${a}\cup{ah^2+bh+c:0\leq h<p} = {0, 1, 2}$ and the cardinality of this set is odd. Is there any example where the cardinality of this set is even and no of QRs $\neq$ no of QNRs? – PTDS Aug 09 '19 at 15:11
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it suffices to consider a=1 b=0 c not 0 consider t from 1 to p-1 and consider the equation h^2+c=(h+t)^2 linear in h and hence 1 solution further it is clear that now we see that 2 different t1,t2 produce the same h solution iff (h+t1)^2=(h+t2)^2 iff h+t1=-(h+t2) or t1=-2h=t2 now choosing over all t we get that half of $\{ah^2+bh+c:1\leq h<p\}$ are squares fufilling the claim.

Hao S
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