0
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There is a list where, using the coordinates of the x points, it was determined whether there are points in the curve

Here's a link

It can be seen that the generator according to the formula y ^ 2 = x ^ 3 + a * x + b determined from the list GPoint = (Gx, Gy) # Generator Point

a= 0

b= 7

p= 115792089237316195423570985008687907853269984665640564039457584007908834671663


y^2 = x^3 + a * x + b # secp256k1

point   (x,y)

point   (1,29896722852569046015560700294576055776214335159245303116488692907525646231534)

point   (2,69211104694897500952317515077652022726490027694212560352756646854116994689233)

point   (3,94471189679404635060807731153122836805497974241028285133722790318709222555876)

point   (4,40508090799132825824753983223610497876805216745196355809233758402754120847507)

point   (5,0)

point   (6,19112057249303445409876026535760519114630369653212530612662492210011362204224)

point   (7,0)

point   (8,91736135629086734185706894124002126994554994840140056297753929940646699135966)

point   (9,0)

point   (10,0)

.......
.......
.......
.......
etc

But I, on the contrary, need to determine by the list through the point y

that is

point (y, x)

Is it possible to do this?

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1
  • 2
    $\begingroup$ This looks like a CTF. What have you done? Hint: it's algebra in finite field $\mathbb F_p$. Many of the algebra techniques you learned to solve equations in the field $\mathbb R$ work, and here $a=0$ will make it easy. When it comes to inverting $z\mapsto z^3$ in this field, Fermat's Little Theorem comes to the rescue. $\endgroup$
    – fgrieu
    Sep 28 '21 at 21:21

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