Timeline for 4900, a particularly square number
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2010 at 2:55 | comment | added | Gerry Myerson | It seems no one has picked up on the question of whether "starting from 1" is necessary. Of course, it is; trivially, every square integer is the sum of (one) consecutive square(s). Slightly less trivially, $5^2=3^2+4^2$, $29^2=20^2+21^2$ (the first two members of an infinite sequence of examples obtainable via Pell). There's also $38^2+39^2+\dots+48^2=143^2$, again the first in an infinite sequence of square sums of 11 consecutive squares, and many, many more. | |
| May 31, 2010 at 14:09 | vote | accept | Michael Hoffman | ||
| May 31, 2010 at 11:05 | answer | added | Roland Bacher | timeline score: 15 | |
| May 31, 2010 at 9:11 | history | edited | Wadim Zudilin |
edited tags
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| May 31, 2010 at 9:10 | vote | accept | Michael Hoffman | ||
| May 31, 2010 at 14:07 | |||||
| May 31, 2010 at 8:55 | answer | added | Charles Matthews | timeline score: 15 | |
| May 31, 2010 at 8:36 | comment | added | Michael Hoffman | Jonas, you could just put that as the answer, that'll get me where I need to be. The CannonballProblem link doesn't help unless you have the papers. | |
| May 31, 2010 at 8:31 | comment | added | Michael Hoffman | Yes, I know that it's the sum from 1 to 24 squared, that was easy, I'm wondering WHY it's the only answer. | |
| May 31, 2010 at 8:30 | history | edited | Michael Hoffman | CC BY-SA 2.5 |
made more clear, hopefully
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| May 31, 2010 at 8:22 | comment | added | J.C. Ottem | $1^2+2^2+...+24^2=70^2=4900.$ | |
| May 31, 2010 at 8:21 | comment | added | Jonas Meyer | math.ubc.ca/~bennett/paper21.pdf | |
| May 31, 2010 at 8:19 | comment | added | KConrad | mathworld.wolfram.com/CannonballProblem.html | |
| May 31, 2010 at 8:15 | comment | added | Jonas Meyer | What is the question? | |
| May 31, 2010 at 8:11 | history | asked | Michael Hoffman | CC BY-SA 2.5 |