Timeline for Can repunits be perfect cubes?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2013 at 13:23 | answer | added | Luara | timeline score: 0 | |
| Sep 10, 2013 at 12:15 | history | edited | Ricardo Andrade |
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| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 25, 2013 at 23:17 | |||||
| May 30, 2013 at 14:24 | vote | accept | Wangt Fei | ||
| May 29, 2013 at 1:11 | answer | added | Noam D. Elkies | timeline score: 18 | |
| May 28, 2013 at 5:30 | vote | accept | Wangt Fei | ||
| May 30, 2013 at 14:24 | |||||
| May 27, 2013 at 22:02 | answer | added | Noam D. Elkies | timeline score: 12 | |
| May 27, 2013 at 17:04 | comment | added | user9072 | @Gerhard Paseman: yes, but the $R_n$ in the question, to which my "this" refers, is base ten, else this would not be equivalent to the given equation. | |
| May 27, 2013 at 16:56 | comment | added | Gerhard Paseman | Quid, in base 18 there is a cube: 324 + 18 + 1 = 343. Gerhard "Ask Me About System Design" Paseman, 2013.05.27 | |
| May 27, 2013 at 16:51 | comment | added | Barry Cipra | @Tom (and Wangt Fei) sorry, you're absolutely right. I made a very stupid mistake. I somehow miscomputed $(10a+1)^3 \equiv30a^2 + 1$ instead of $30a+1$. | |
| May 27, 2013 at 16:17 | answer | added | user9072 | timeline score: 10 | |
| May 27, 2013 at 15:59 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added 35 characters in body; edited title
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| May 27, 2013 at 15:48 | comment | added | Tom De Medts | @Barry: This elementary method doesn't work. In fact, it's not too hard to see that for every positive integer $d$, there exists an integer whose cube ends in at least $d$ digits $1$. (And this number is unique modulo $10^{d+1}$.) | |
| May 27, 2013 at 15:39 | comment | added | user9072 | In case it is of interest, it is even known this cannot be a perfect power (not just not a cube), proved by Bugeaud and Mignotte in "Sur l'équation diophantienne $(x^n - 1)/(x-1)= y^q$, II" (see Thm 5). | |
| May 27, 2013 at 15:34 | comment | added | Wangt Fei | I know that Rn≡n(mod 9), and if Rn is a cubic number then n≡0, ±1(mod 9). The left appears difficult. | |
| May 27, 2013 at 15:21 | comment | added | Barry Cipra | I think this question is better suited for math.stackexchange. It looks like a straightforward exercise in elementary number theory. (Hint: Solve $b^3\equiv1\mod10$ for $b$ and then $(10a+b)^3\equiv11\mod100$ for $a$.) | |
| May 27, 2013 at 15:10 | history | edited | Wangt Fei | CC BY-SA 3.0 |
added 1 characters in body; deleted 1 characters in body
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| May 27, 2013 at 15:04 | history | asked | Wangt Fei | CC BY-SA 3.0 |