2

I've been trying to understand the issue and repeat Willem Hengeveld's answer, but I think I'm failing at getting correct script for hash.

I'm looking for this transaction, my scripts are:

010000000396773312284612e55b124e19b885038f625ce18f99711c9138d22c391a2abc86000000001976a914fa6b5cff131a62d212f5c99d8bfaf35a9299ad5288acffffffff2f4fcd2541be1e286b7eee1a24bf0b9d6a6333c05ed7955d083f0778e2a23efc0000000000ffffffff40dc504498fa0ff41f2a6d882b640a17a4fbdb96a34ff28ad7f83c85fd78521e0000000000ffffffff0240420f00000000001976a91406f1b66ffe49df7fce684df16c62f59dc9adbd3f88ac66732c00000000001976a914fa6b5cff131a62d212f5c99d8bfaf35a9299ad5288ac0000000001000000

010000000396773312284612e55b124e19b885038f625ce18f99711c9138d22c391a2abc860000000000ffffffff2f4fcd2541be1e286b7eee1a24bf0b9d6a6333c05ed7955d083f0778e2a23efc000000001976a914fa6b5cff131a62d212f5c99d8bfaf35a9299ad5288acffffffff40dc504498fa0ff41f2a6d882b640a17a4fbdb96a34ff28ad7f83c85fd78521e0000000000ffffffff0240420f00000000001976a91406f1b66ffe49df7fce684df16c62f59dc9adbd3f88ac66732c00000000001976a914fa6b5cff131a62d212f5c99d8bfaf35a9299ad5288ac0000000001000000

And double SHA256 are:

z1 = 1a374484e71c12135cda76c1261801e9f8fc27108e3f5e439679ef13eb300eae

z2 = e366e7bc7d5c12808b88e1b89c27436ac9cb3ae482dd044faeb7fab4654cbeac

Even applying +/- for s1,s2 I'm getting those 4 private keys:

68d4ee844087fc39a62269c41a2c7d521966aa9687a596ab9207e997ddee0ba3
5d201313d68f45cfd5b187584efe4e5d9579a571a0876a3e1fb649ed91942897
a2dfecec2970ba302a4e78a7b101b1a1253537750ec135fda01c149f3ea218aa
972b117bbf7803c659dd963be5d382aca148325027a309902dca74f4f248359e

Which do not corresponded to that public key. What I'm doing wrong?

1

You are not writing what other values you have used. The shasha you calculated is correct.

pubkey = 044235f67fa40e91b4bc8c940ffdefc8f4f74c4e8436784e470115f0d48af40292922cba638ed6028ecb8b23280111f778cf2d200bbff4dabaf53cf821ce135261
r = 6a93787fbfef76b95d2f4d7632b841459d440afaaa0873608436efcd3fbe3e54
s1 = e939c2360b5dcdb64c2bc8d5ffad06a7cdfde79f29ec967a5d5d07cdec142815
s2 = 0df72a51154081325f81c87eb220387833aac888dcc4800b78d8423f59b1a711
m1 = 1a374484e71c12135cda76c1261801e9f8fc27108e3f5e439679ef13eb300eae
m2 = e366e7bc7d5c12808b88e1b89c27436ac9cb3ae482dd044faeb7fab4654cbeac

then ( see https://ideone.com/M1xWFJ for a python script demonstrating)

k = 6e3469cb1dec3ce994dfc5c88bb53971fe513749727bdfa4a44a38f294008136
x = 1930a0cb6fe514b9ab03c652a61ac53b2c7ee6db417543de782503e690fab966
2
  • My s2 was a bit wrong by one bit, thank you for pointing out it to me :)
    – nateless
    May 25 '16 at 20:56
  • How to get m1 and m2 Thank you for your help Dec 25 '20 at 1:45

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