Let $E$ be the elliptic curve $y^2=x^3+1$ and $p \equiv 1 \pmod{3}$ a prime. Computing the number of points mod $p$ of $E$ using the naive method gives: $$ \#E(\mathbb F_p) = 1+ \sum_{x=0}^{p-1} \left( 1 + \left(\frac{x^3+1}{p}\right)\right) \equiv p+1 + \sum_{x=0}^{p-1}(x^3+1)^{\frac{p-1}2}\pmod{p}$$ developing the right hand side and simplifying there is only one term of the sum surviving and we get: $$ \#E(\mathbb F_p) \equiv p+1-\binom{\frac{p-1}2}{\frac{p-1}3}\pmod{p}$$ Hasse theorem now implies that if $p$ is large enough (say $p>16$) then the minimal residue of $$ \binom{\frac{p-1}2}{\frac{p-1}3}\pmod{p}$$ is smaller than $2\sqrt{p}$. As this last result seems completely unrelated to elliptic curves my question is if there is a more direct proof of this fact not using Hasse's theorem.

Using instead the elliptic curve $y^2 = x^3+x$ and a prime $p \equiv 1 \pmod{4}$ one finds the similar result that the least minimal residue of $$ \binom{\frac{p-1}2}{\frac{p-1}4}\pmod{p} $$ is also bounded in absolute value by $2\sqrt{p}$.

My motivation is simple curiosity, I fell on it playing with some examples and now I can't take it out of my head.