Quote from: BitcoinADAB on July 28, 2021, 12:27:14 AM

Point 1 (x1, y1)
x1 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y1 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc

Point 2 (x2, y2)
x2 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y2 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc

Point 3 (x3, y3)
x3 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y3 = 0xeb3c392e5ac716a0cb40fa08e2616f47459e6a1cc0f2922836896a1ce5f631cc

Point 4 (x4, y4)
x4 = 0x991eb8eb2e45b4bc9c71bc9a022832e712a8dc1b2db62bd7456e49b2d9f7dac8
y4 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

Point 5 (x5, y5)
x5 = 0xa673e97568057fb5f41c35d6ed6c88ef97510d71222b3686ef892f4ccc2af536
y5 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

Point 6 (x6, y6)
x6 = 0xc06d5d9f69b4cb8d6f720d8f106b442956061673b01e9da1cb0886fe59dd2860
y6 = 0x14c3c6d1a538e95f34bf05f71d9e90b8ba6195e33f0d6dd7c97695e21a09ca63

I have a question.
Can you tell me the private key corresponding to points 1 to 6?
Then I can test these public keys in a different way.

private key : 0xa
public key : 0xa0434d9e47f3c86235477c7b1ae6ae5d3442d49b1943c2b752a68e2a47e247c7, 0x893aba425419bc27a3b6c7e693a24c696f794c2ed877a1593cbee53b037368d7

private key : 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364137
public key : 0xa0434d9e47f3c86235477c7b1ae6ae5d3442d49b1943c2b752a68e2a47e247c7, 0x76c545bdabe643d85c4938196c5db3969086b3d127885ea6c3411ac3fc8c9358

For example
Points 1 and 4 can be easily calculated,
Is there a formula for calculating the private key of point 2 and 3?