Timeline for How did Cole factor $2^{67}-1$ in 1903?
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2019 at 8:57 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
edited title
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| Mar 10, 2019 at 23:14 | answer | added | Federico Poloni | timeline score: 30 | |
| Feb 12, 2019 at 17:36 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
fix spacing of non-punctuating commas in TeX
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| S Feb 12, 2019 at 16:01 | history | suggested | bruno | CC BY-SA 4.0 |
Uses punctuation to make it a little bit more readable
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| Feb 12, 2019 at 15:25 | review | Suggested edits | |||
| S Feb 12, 2019 at 16:01 | |||||
| May 23, 2015 at 18:45 | comment | added | Sylvain JULIEN | Perhaps mathematicians of these times relied more on their own intuition than we do...Rigor makes your path secure and accurate but intuition makes you walk way faster. | |
| May 23, 2015 at 14:08 | answer | added | Stefan Kohl♦ | timeline score: 14 | |
| May 22, 2015 at 21:24 | answer | added | David E Speyer | timeline score: 44 | |
| May 22, 2015 at 16:04 | vote | accept | David E Speyer | ||
| May 22, 2015 at 15:59 | comment | added | GH from MO | @AnthonyQuas: Let $p\mid 2^{67}-1$ be a prime. Look at the order $d$ of $2$ modulo $p$. We have that $d\mid 67$ and $d\mid p-1$, but $d>1$. So $d=67$, and hence $67\mid p-1$. | |
| May 22, 2015 at 15:59 | comment | added | David E Speyer | Nope, issue with the factor tables sill exists in $1903$. @AnthonyQuas Since $2^{67} \equiv 1 \bmod N$, and $GCD(2-1, N)=1$, for any prime $p$ dividing $N$, there is a nontrivial $67$-th root of unity modulo $p$, and that forces $p \equiv 1 \bmod 67$. | |
| May 22, 2015 at 15:58 | answer | added | user9072 | timeline score: 66 | |
| May 22, 2015 at 15:57 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 43 characters in body
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| May 22, 2015 at 15:57 | comment | added | Anthony Quas | Highly ignorant question: I see that if $2^{67}-1$ is written as $pq$, then $pq\equiv 1\pmod {67}$. Why are $p$ and $q$ individually congruent to 1? | |
| May 22, 2015 at 15:57 | answer | added | Gerhard Paseman | timeline score: 8 | |
| May 22, 2015 at 15:54 | comment | added | GH from MO | Perhaps he used a mechanic calculator for his divisions: en.wikipedia.org/wiki/Mechanical_calculator | |
| May 22, 2015 at 15:53 | comment | added | David E Speyer | @SamHopkins Whoa, how did I get that wrong? 1876 is the date that Lucas invents the Lucas-Lehmer test to show that $N$ is composite. Fixing the title, thanks. And now back to Dickson to see which prime tables existed... | |
| May 22, 2015 at 15:52 | history | edited | David E Speyer | CC BY-SA 3.0 |
edited title
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| May 22, 2015 at 15:51 | comment | added | Sam Hopkins | Wikipedia (en.m.wikipedia.org/wiki/Frank_Nelson_Cole) suggests Cole factored this number in 1903 (or perhaps 1900-1903), if that makes any difference in terms of tools available at the time... | |
| May 22, 2015 at 15:50 | comment | added | David E Speyer | Actually, it takes us down to $2$ a minute, since the odd numbers are already not in our table of primes. | |
| May 22, 2015 at 15:42 | comment | added | David E Speyer | Although the issue about the prime tables not existing still bothers me... | |
| May 22, 2015 at 15:40 | comment | added | The Masked Avenger | Oh, and remember to divide by 2-1 first. | |
| May 22, 2015 at 15:40 | comment | added | David E Speyer | Ah, thank you. And that takes us down to a little under $1$ a minute, which isn't crazy, though still hard. | |
| May 22, 2015 at 15:40 | comment | added | GH from MO | @TheMaskedAvenger: I beat you by 6 seconds! :-) | |
| May 22, 2015 at 15:39 | comment | added | The Masked Avenger | Fermat-Lagrange, which says that p is 134k+1 for some integer k. | |
| May 22, 2015 at 15:39 | comment | added | GH from MO | Well, $67$ is a prime, so all prime factors of $2^{67}-1$ are congruent to $1$ mod $134$. | |
| May 22, 2015 at 15:33 | history | asked | David E Speyer | CC BY-SA 3.0 |