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Author Topic: Mini-puzzle #4  (Read 1035 times)
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kTimesG
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January 15, 2025, 03:02:19 PM
Last edit: January 15, 2025, 03:13:08 PM by kTimesG
 #21

WTF are you all talking about, the points are not next to each other, they are distanced by x*BETA.

Whatever point you choose as a generator, you can always solve the equations:

k1*lam = k1 + 1
k2*lam = k2 - 1
k3*lam*lam = k3 + 1
k4*lam*lam = k4 - 1
k5 = -k5 + 1 (k = 1/2)
k6 = -k6 - 1 (k = -1/2)

and reach the same "conclusion" that two adjacent keys share the same Y, but they are in no way next to each other on the curve. Only the private key scalar are consecutive.

I don't think two points exist actually next to each other and separated by a scalar of 1.


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January 15, 2025, 03:12:02 PM
 #22

WTF are you all talking about, the points are not next to each other, they are distanced by x*BETA.

Whatever point you choose as a generator, you can always solve the equations:

k1*lam = k1 + 1
k2*lam = k2 - 1
k3*lam*lam = k3 + 1
k4*lam*lam = k4 - 1

and reach the same "conclusion" that two adjacent keys share the same Y, but they are in no way next to each other on the curve. Only the private key scalar are consecutive.
I don't think two points exist actually next to each other and separated by a scalar of 1.

Take some point Point1. Then take Point2 = Point1 + G. I call Point2 as next point after Point1. Don't you agree?

I've solved #120, #125, #130. How: https://github.com/RetiredC
kTimesG
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January 15, 2025, 03:16:30 PM
 #23

WTF are you all talking about, the points are not next to each other, they are distanced by x*BETA.

Whatever point you choose as a generator, you can always solve the equations:

k1*lam = k1 + 1
k2*lam = k2 - 1
k3*lam*lam = k3 + 1
k4*lam*lam = k4 - 1

and reach the same "conclusion" that two adjacent keys share the same Y, but they are in no way next to each other on the curve. Only the private key scalar are consecutive.
I don't think two points exist actually next to each other and separated by a scalar of 1.

Take some point Point1. Then take Point2 = Point1 + G. I call Point2 as next point after Point1. Don't you agree?

No. They are simply consecutive keys, not points next to each other. How is you example different from having k = -1/2 and k = 1/2 ? [-1/2]G and [1/2]G are consecutive points also? Or are they actually opposite points?

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January 15, 2025, 03:31:52 PM
 #24

WTF are you all talking about, the points are not next to each other, they are distanced by x*BETA.

Whatever point you choose as a generator, you can always solve the equations:

k1*lam = k1 + 1
k2*lam = k2 - 1
k3*lam*lam = k3 + 1
k4*lam*lam = k4 - 1

and reach the same "conclusion" that two adjacent keys share the same Y, but they are in no way next to each other on the curve. Only the private key scalar are consecutive.
I don't think two points exist actually next to each other and separated by a scalar of 1.

Take some point Point1. Then take Point2 = Point1 + G. I call Point2 as next point after Point1. Don't you agree?

No. They are simply consecutive keys, not points next to each other. How is you example different from having k = -1/2 and k = 1/2 ? [-1/2]G and [1/2]G are consecutive points also? Or are they actually opposite points?

Having consecutive keys means having consecutive points. If you don't agree - no problem, I'm fine with that  Grin

I've solved #120, #125, #130. How: https://github.com/RetiredC
kTimesG
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January 15, 2025, 03:50:42 PM
Last edit: January 15, 2025, 04:05:03 PM by kTimesG
 #25

Another explanation:
It's a special point, if you apply endomorphism for it, you will get previous point (Point - G).
It's funny to have two points next to each other with same Y, isn't it?  Smiley
And since you know the distance between these two points, you can calculate the private key easily.

It was all about the wording, you made it sound as if there are two points next to each other. They are actually private keys next to each other. And the same goes for the distance - it is the distance between the private keys. Anyway, there are at least four keys that have the property which is the basis of your puzzle (same Y, adjacent keys), not just a single one. A better question would be: shouldn't there actually be 6 of them? What happened to lambda^3 = 1? Hmm.

LE - I think there are maaaany more other solutions. For example:

(k/2)*lam = (k/2) + 1
k = 26024669624720066173548254508404295800636635574530772464662847532876721171439

kG.y = (k+1)G.y =
0x99dabfaa9e4dac1be74714258a4f862f2694e9d6e6b74f9fe979a7ebeec465fc

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RetiredCoder (OP)
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January 15, 2025, 04:04:42 PM
 #26

Another explanation:
It's a special point, if you apply endomorphism for it, you will get previous point (Point - G).
It's funny to have two points next to each other with same Y, isn't it?  Smiley
And since you know the distance between these two points, you can calculate the private key easily.

It was all about the wording, you made it sound as if there are two points next to each other. They are actually private keys next to each other. And the same goes for the distance - it is the distance between the private keys. Anyway, there are at least four keys that have the property which is the basis of your puzzle (same Y, adjacent keys), not just a single one. A better question would be: shouldn't there actually be 6 of them? What happened to lambda^3 = 1? Hmm.

You are much more of a nerd than I am  Cheesy

I've solved #120, #125, #130. How: https://github.com/RetiredC
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