During my research, I came across this question.
Let $p>11$ a prime number with $a=\text{card}\{(x,y) \in \mathbb Z/ p \mathbb Z: y^2=x^3+1\}$ and $b=\dfrac 1 {((p-1)/2)! \times ((p-1)/3)! \times ((p-1)/6)!} \mod p$ when $p \mod 3=1$.
Is it true that if $p \mod 3=1$ then $a \in \{b,p-b, \min\{b,p-b\}+p\}$ else $a=p$?
Elliptic curves of the form $y^2=x^3+b$ were studied by Schrutka in the article Ein Beweis für die Zerlegbarkeit der Primzahlen von der Form $6n +1$ in ein einfaches und ein dreifaches Quadrat. He used only elementary arguments. In particular he has proven congruences for involved numbers.
Schrutka used sums in the form $$\sum_{x=1}^{p-1}\left(\frac{x^4+n x}{p}\right),$$ but they are essentually the same as the sums $$\sum_{x=1}^{p-1}\left(\frac{x^3+b}{p}\right),$$ because $$\sum_{x=1}^{p-1}\left(\frac{x^4+n x}{p}\right)=\sum_{x=1}^{p-1}\left(\frac{x^{-4}+n x^{-1}}{p}\right)=\sum_{x=1}^{p-1}\left(\frac{1+n x^3}{p}\right)=\left(\frac{n}{p}\right) \sum_{x=0}^{p-1}\left(\frac{x^3+n^{-1}}{p}\right).$$
Related questions are
Direct proof of special case of Hasse's theorem for elliptic curves and
Congruence for the number of points in the elliptic curve $y^2=x^3+b\pmod p.$
