Timeline for Proof that user public key corresponds the curve equation (secp256r1)
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2023 at 18:44 | comment | added | kelalaka | @unalignedmemoryaccess it is by definition. Dou want to see longer format, see it here neuromancer.sk/std/nist/P-256 | |
| Dec 21, 2023 at 16:25 | comment | added | unalignedmemoryaccess |
Why is there -3 in the equation for a256?
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| Nov 8, 2020 at 22:18 | history | edited | kelalaka | CC BY-SA 4.0 |
little corrections.
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| Oct 26, 2020 at 16:11 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Oct 26, 2020 at 10:03 | comment | added | kelalaka | A little explanation added to the answer. | |
| Oct 26, 2020 at 10:02 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Oct 26, 2020 at 9:50 | comment | added | Steven | Alright, thank you very much for your time and help. Things became much clearer! :) | |
| Oct 26, 2020 at 9:33 | comment | added | kelalaka |
Because the curve is defined over a finite field where p256 on the code for secp256r1. This means that all the points have coordinate in the field and arithmetic performed according to it. In simple terms, you are working at $Z_p$.
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| Oct 26, 2020 at 9:15 | comment | added | Steven |
Thank you, it became way more clear now! :) I still missunderstand something. Why do you calculate modulo the field? I've tested the online IDE aswell and tried if px^3+a256*px+b256 equals to py^2 but it doesn't, so I think my whole understanding is wrong... I'm really sorry for being so slow, but even with researching a lot I don't seem to understand it and that's really frustrating.
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| Oct 26, 2020 at 9:08 | history | edited | kelalaka | CC BY-SA 4.0 |
deleted 3 characters in body
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| Oct 26, 2020 at 9:02 | comment | added | kelalaka | I've simplified the code. now, it just calculates the arithmetic, | |
| Oct 26, 2020 at 9:01 | history | edited | kelalaka | CC BY-SA 4.0 |
removed the unnecessary code.
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| Oct 26, 2020 at 8:56 | comment | added | kelalaka | Well, I don't know what do you mean by hand, however, the last 3 lines you need with $a$,$b$, and $p$. For the, you need a big number library where SageMath is helpful. | |
| Oct 26, 2020 at 8:53 | vote | accept | Steven | ||
| Oct 26, 2020 at 8:53 | comment | added | Steven | Wow, thanks a lot for the effort! :) Unfortunately I'm supposed to figure it out "by hand". Do you know if the approach I've taken up there is wrong? All my research so far showed that this seems to be the most common method to check it out. But due to your confirmation I do atleast know that my calculation is wrong and I can be sure that my trial and error is not for nothing. :'D P.S: Thanks for the hint with the square root, I didn't think about it properly. | |
| Oct 26, 2020 at 8:47 | history | answered | kelalaka | CC BY-SA 4.0 |