Quote from: Legends_Never_Die on Today at 01:29:47 PM

Ok, I will give you this, when ever you managed to give an accurate and correct answer, I will believe that point by point multiplication is also possible, not saying there is no way, of course there is, we only need to know at least a mutual factor, a distinct divisor of a point to calculate the result of point to point multiplication.

Tell me if it's possible to have 2+2= 5, in the future when we advance in science etc?

^{p = 6
pubkey = bitcoi[Suspicious link removed]ivkey_to_pubkey(p)
P1 = Point(SECP256k1.curve,
pubkey[0],
pubkey[1])

p = 8
pubkey = bitcoi[Suspicious link removed]ivkey_to_pubkey(p)
P2 = Point(SECP256k1.curve,
pubkey[0],
pubkey[1])

p = 48
pubkey = bitcoi[Suspicious link removed]ivkey_to_pubkey(p)
P3 = Point(SECP256k1.curve,
pubkey[0],
pubkey[1])}

As you can see these are 3 points. 3rd one is the multiplication of first two.
Now try this:

from sympy import symbols, Eq, solve
_{a, b = symbols('a b')

# Points
point1 = (53957576663012291606402345341061437133522758407718089353314528343643821967563,
98386217607324929854432842186271083758341411730506808463586570492533445740059)
point2 = (72488970228380509287422715226575535698893157273063074627791787432852706183111,
62070622898698443831883535403436258712770888294397026493185421712108624767191)
result_point = (27014530906007957169531703425481274912103442539779119437726046076907210423569,
70052204206988426097199854494650803460514936258189950471015657998938191627469)

# Setting up the equations for linear combination
eq1 = Eq(a * point1[0] + b * point2[0], result_point[0])
eq2 = Eq(a * point1[1] + b * point2[1], result_point[1])

# Solving the equations
solution = solve((eq1, eq2), (a, b))
print(solution)}

When you do another multiplication a and b will get change. Of course this won't be the formula it will be something like:
_{m = ((c * point1.x() * point2.x()) + curve.a()) * \
inverse_mod(2 * point1.y() * point2.y(), curve.p())
x3 = (m * m - 2 * point1.x() * point2.x()) % curve.p()
y3 = (m * (point1.x() * point2.x() - x3) - point1.y() * point2.y()) % curve.p()}

This is not the formula of course, I just copied and enhanced from point doubling, but something like this will give the point multiplication result for all cases. I hope you got my idea. Rather than discussing the possibility of some operations, we must start to think about realizing it because I can say that honestly I clearly see that can be done. And point exponent will be most probably the same as scalar multiplication, just we won't use point_add, we will use point_multiply by itself.