Timeline for On choosing the correct square root of $g^{4n}$ modulo primes
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2022 at 7:08 | comment | added | joro | @BenSmith Thanks, you have valid concerns about periodicity and "exponent overflow". Edited trying to fix the definition by assuming $ 1 < n < (p-3)/4$. | |
| Dec 18, 2022 at 7:05 | history | edited | joro | CC BY-SA 4.0 |
addressed comments
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| Dec 16, 2022 at 11:00 | comment | added | Ben Smith | (1) is straightforward, and (2) and (3) are consequences of $g$ being a generator (so $g^{(p-1)/2} = -1$). | |
| Dec 16, 2022 at 10:42 | comment | added | Ben Smith | I'm not saying anything about (1) and (2), I'm saying that the definition I quoted is incoherent. | |
| Dec 16, 2022 at 1:02 | review | Close votes | |||
| Dec 22, 2022 at 3:03 | |||||
| Dec 16, 2022 at 0:43 | comment | added | Max Alekseyev | The questions are too vague. Can you make them more specific? | |
| Dec 15, 2022 at 17:13 | comment | added | joro | @BenSmith Thanks. Are you saying that (1) and (2) don't hold? Experimentally for p=31 they hold up to n=31^2. | |
| Dec 15, 2022 at 15:35 | comment | added | Ben Smith | The definition "For all positive integers $n$, the correct square root modulo $p$ of $g^{2n}$ is $g^n$ is problematic: $g^{2n} \equiv g^{2(n + (p-1)/2)}$ for any $n$, but $g^n \equiv -g^{n+(p-1)/2}$. | |
| Dec 15, 2022 at 15:20 | comment | added | joro | @ChristopheLeuridan Thanks. I edited with clarification about your comment. | |
| Dec 15, 2022 at 15:19 | history | edited | joro | CC BY-SA 4.0 |
added 80 characters in body
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| Dec 15, 2022 at 15:07 | comment | added | Christophe Leuridan | What is $n$? Any integer? An integer in $[0,(p-3)/4]$? What do you mean by the correct square root? | |
| Dec 15, 2022 at 13:01 | history | asked | joro | CC BY-SA 4.0 |