Quadratic Residues in $\{a\}\cup\{ah^2+bh+c:0\leq h<p\}$

Question

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 $p$?

Answer 1

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.

Answer 2

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