Point-count criterion for y squared equals x cubed plus k x ($\#E({\mathbb{F}}_p)=p+1⟺p\equiv 3(\mathrm{mod}4)$)

For odd $p$ and $k\not\equiv 0(\mathrm{mod}p)$, take $E:y^2=x^3+kx$. If $p\equiv 3(\mathrm{mod}4)$, pairing $x$ with $-x$ cancels the Legendre symbols. If $p\equiv 1(\mathrm{mod}4)$, the coefficient formula for trace of Frobenius modulo p givesThe binomial coefficient is nonzero because every factorial involved is shorter than $p$. Thus the Trace of Frobenius is not zero in the second case.