Let, $E:=y^2=x^3 + Ax + B$ an elliptic curve, 2 points $P, -Q, Q$ on $E$ such that $2P=Q$, we can write
$$y^2-(x^3 + Ax + B) = (x - e_1)(x - e_2)^2=0$$ Where, $e_1=x(Q), e_2=x(P)$ a double root. So, $y=m(x-x(Q))+y(Q)$, here $y$ is the line that goes thorough $P, Q$, and $m$ is the slope.
$$ (x - e_1)(x^2 - 2xe_2+e_2^2)$$ $$=(x^3 - 2x^2e_2+xe_2^2)-(x^2e_1 - 2xe_1e_2+e_1e_2^2)$$ $$=x^3 - 2x^2e_2+xe_2^2-x^2e_1 + 2xe_1e_2-e_1e_2^2$$ $$=x^3 + x^2(-1)(2e_2+e_1) +x(e_2^2+ 2e_1e_2)+(-1)e_1e_2^2$$
From $y^2-x^3 + Ax + B =0$ we get coefficient of $x^2$ is $(-1)m^2$, equating coefficient we find, $m^2=(2e_2+e_1)$.
Problem is , if I simplify both side and then equate the coefficient, then, I get correct coefficient of $x^2$ but not for $x$ (if we plug in value of $x, y$ coordinate of points $P, Q$, the coefficients of $x^2$ satisfies only, not $x$).
What is the problem?
Edit:
$x(P), y(P)$ are the co-ordinate with respect to $x$ and $y$ axis respectively, for point $P$.
$A, B$ are integer and $(x,y)$ is point on $E.$
I have the below identities:
$$2e_2+e_1=m^2,$$
$$ e_2(e_2+2e_1)=A+2m(mx(Q)-y(Q)),$$
$$e_1e_2^2=B-(y(Q)-mx(Q))^2$$.
but it does not work for last 2, try this : Let $E$ be the elliptic curve $y^2 = x^3 − 25x$, $P = (−4,6), 2P = (\frac{1681}{ 144} ,\frac{ −62279}{ 1728 }), m = \frac{23} {12}.$
Let $m'$ be the slope of the line going through $P$ and $-Q$, ie the tangent to the elliptic curve at $P$.
What we know is that the factorization $x^3+Ax+B-(m'(x-x(-Q))+y(-Q))^2=(x-x(-Q))(x-x(P))^2=(x-e_1)(x^2-2e_2x+e_2^2)$ holds (as $P$,$P$,$-Q$ are collinear).
So, $x^3-m'^2x^2+x(A+2m'^2x(Q)+2m'y(Q))+B-(y(Q)+m'x(Q))^2=x^3-(2e_2+e_1)x^2+e_2(e_2+2e_1)x-e_1e_2^2$.
Therefore $2e_2+e_1=m'^2$, $e_2(e_2+2e_1)=A+2m'(m'x(Q)+y(Q))$, $e_1e_2^2=B-(y(Q)+m'x(Q))^2$. Do these identities work?