Is there an efficient formula for computing point halving on elliptic curves in $\operatorname{char}F=p\ne2$? Consider the elliptic curve in $F_p$ and let $n$ be the order of the group.

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*Suppose the curve is the same as bitcoin: $y^2 = x^3 + 7$, I have a point $P=(x,y)$，how to compute $\frac{1}{2}P$.


*Suppose $P=aG$ for some base point $G$(we know the points $P,G$, but the coefficient $a$)，then is there efficient method to compute $\lfloor\frac{a}{2}\rfloor$G
 A: Case $P = \lfloor \frac a 2 \rfloor G$;
As said in the comments if possible this can help to break the DLog.
Case $P = \frac a 2 G$;
We may expect that it has two solutions $$P =\frac{x}{2} \cdot G =\frac{x + q}{2} \cdot G$$, however, the base point $G$ of the bitcoin curve has odd order. Therefore
int("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141", 16)
115792089237316195423570985008687907852837564279074904382605163141518161494337

Therefore if there exists half of a point then it is unique.

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*The derivative of the equation $y^2 = x^3 + 7$
$\lambda = \dfrac{3x^{2}}   
    {2(x^{3}  + 7)^{1/2}} \tag{1} \label{1}$


*Write down the equation for the tangent line to $E$ at G:

$$y-p_y = \lambda(x-p_1) => y = \lambda(x - p_x) + p_y \tag{2}$$


*Write the equation for the intersection of the tangent line with E:

$$y^{2} = (\lambda(x - p_x) + p_y)^{2} = x^{3} + 7 \tag{3}\label{3}$$


*Write down $\lambda$ in terms of $p_x$:
If we extend the equation \ref{3} we will get a monic cubic polynomial, so the coefficient of $x^2$ will be the minus the sum of the roots. Therefore, $p_x$ will be double and $g_x$ will be a single root.  $$(\lambda (x - p_x) + p_y)^{2} = x^{3} + 7$$ $$\lambda^2 (x - p_x)^2 + 2 \lambda (x - p_x) p_y + p_y = x^{3} + 7$$ as we can see the term for $x^2$ is $\lambda$. Since the roots are the intersection point then $p_x$ will be a double point and $g_x$ will be a single point, so $2 p_x + q_x = \lambda^2$ $$\lambda^2 = 2 p_x + q_x  \label{4}\tag{4}$$


*Form equation using equation \ref{4} and and replacing $x$ with $p_x$ in the equation \ref{1};
$$\lambda^{2} = 2p_x + q_x  =  \dfrac{3p_x^{4}}{4(p_x^{3}  + 7)} \tag{5} \label{5}$$ Simplify this equation and move everything over to the left-hand side will yield $$ (2p_x + q_x)(4(p_x^{3}  + 7))  =  3p_x^{4} $$ $$ 5 p_x^4 + 4 q_x p_x^3 + 28 q_x + 56 p_x = 0$$ a quartic polynomial in $p_x$. The roots of this polynomial are the x-coordinates of the half points of $G$. We are in the odd case, the factors of the quartic polynomial will be a linear equation and an irreducible resolvent cubic.
