I found on secp256k1 that if 4 points have same doubling slope their private keys always gives 0
k1+k2+k3+k4=0
On smaller curve
P=967 N=907
β (cube root of 1): 142
λ (eigenvalue at P): 522
1G (418, 442)
If I have 1st private key I can calculate 1st column.... and I can calculate all slopes in 1 set of 24 points but I did not find the way to if I have k1 to calculate k2 k3 k4
Also on the big curve I found those sets
f(x) = 9x^4 - 4slope^2x^3 - 28slope^2 = 0
On this way I generated sets of 4 points 48 sets on this curve p=967
I tried other curves with Sagemath
p=1000003 - Number of points: 999006 Number of sets: 41886
p=1001023 - Points: 999066 - Sets:41466
p=1001527 - Points: 1000158 - Sets:41580
p=20000077 - Points: 19992876 - Sets:833040
Finally I tried to search those sets on the big curve. I started with slope 1 and I was searching if current slope will give 4 solutions for that
This is the SageMath code I was using
https://justpaste.it/k055band I got these Slopes and X values
Slope 5:
e5493c5e5ab1e25795a4f01f362ebc855d3c2611100bb0c24b401d5846f28f93
95b49d5352c6d6893771e5524aced66e9c163010d1cf0dcad80a77d786fc4681
91bfb02411f3bbb0e380f649f876e8692acbb596452df00277c846b1191017f5
817b59b8797719a732f66d2814c4683114c58280bc858a53f32607aac3abb2ec
--------------------
Slope 16:
ff405d7729ed52aa591f7be3f59048c4d506fc28ff83f1b78c84f34c9ec19e50
a1f7ff943ae98653ce8e8b90ea104f1f9a669548bf0208a595810c0d526cb547
2fe5f2ec5aed03af8aae566b6a792fdf5ae94630ba878d2a0206ec505e7cd908
126fe8ebce7506e0868730589974711fc3e20bebbfd606b1bf814d375aff7748
--------------------
Slope 47:
d8a0a502488cf53fe88a7b53b3b86021c151e59ef8df9f0f6fe54b821cef7ac6
6ced29459069a670d859aec16c1e85d426d1eaf84a1a14f8beab86e85934b74b
5b260613d299e1e19673ccf072afac0b94e843b73f24fe1f425f1e2ca885df5b
42da6487e2a865fbe18b97335107a6e2112ccf3fb6c4dc11729e484a8c0095df
--------------------
Slope 54:
f9c2e48f573115eddcbc132f9cf5979ac292e3918ecdec4458589501788f5b90
caeb00411e5011e845e8063c58749613f6e51a083896b935cab902a50815f9c2
217212301587cca0fb1387de3db81428cb7b5e95567786fee35fc43327ea2046
19e008ff74f70b88e2485eb5ccddbe287b0ca3d0e223d386f98ea424577087d6
--------------------
Slope 110:
f494b668f9c24a74dcced225e670cd6d9b343b9d8175354123281ab73564f74a
a3c5694b0a6cce2d90e5cfdc9f3bd9a5bad9fab3a0685ece8a5bd958802b979c
35fbd7fee43dafc765aeb3abe77c60fcaba79fbebd7d5b5ffddd70c9acc27109
15384130a5cc1b246580388a766530d38c830d7e59889ec9382cd4084857b888
--------------------
About 16% of all points on the curve have this property
Of course if you know 1 slope then you know p-slope so in the case of any kind of scanning we can just save the slope and we have info for 8 points
Do you think that those points can be used for solving secp256k1? What do you think about all of this?
