Timeline for Rachinsky quintets
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2021 at 9:03 | comment | added | Ben McKay | My wife's grandmother used to tell me how she and her siblings made those shoes for themselves out of reeds. | |
| Jul 1, 2017 at 2:42 | answer | added | Wlod AA | timeline score: 3 | |
| Jun 30, 2017 at 12:37 | comment | added | Wlod AA | Whenever $\ a^2+b^2=c^2,\ $ we get: $$ a^2 + b^2 + c^2\ =\ (a-b)^2 + (a+b)^2 $$ Enjoy! | |
| S Jun 30, 2017 at 11:10 | comment | added | user111753 | We can now consider the geometric puzzle of what is the minimum number of connected blocks that one can dissect the squares on one side and reassemble into the squares on the other side and consider the complexity of the problem. It is not clear to us what a greedy algorithm should be so perhaps it may be easy to prove it is NP hard. | |
| S Jun 30, 2017 at 11:10 | comment | added | user111753 | This is not an answer but I want to post an observation. This quintet is nice because 365 is a number everyone is familiar with and $(10,11,12:13,14)$ is contiguous, but wait so is $(3,4:5)$. May be there are other such contiguous sum of $m+1$ squares which is also sum of $m$ squares. If we let the first term be $a$, we get the quadratic $(a+m)(a-(2m^2+m))$, so there is a unique solution for every $m$ starting with $a=2m^2+m$. The next one for $m=3$ is $(21,22,23,24;25,26,27)$. (continued) | |
| Jun 16, 2017 at 6:12 | history | edited | Zurab Silagadze | CC BY-SA 3.0 |
Broken URL link fixed
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| Jan 17, 2014 at 16:28 | history | edited | KConrad | CC BY-SA 3.0 |
edited body
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| Jan 17, 2014 at 12:33 | comment | added | Zurab Silagadze | @KConrad: Спасибо! Википедия ввел в заблуждение: en.wikipedia.org/wiki/Nikolay_Bogdanov-Belsky | |
| Jan 12, 2014 at 1:07 | comment | added | José Hdz. Stgo. | Just to add that in Yakov Perelman's "Algebra can be fun", one could find the following grade-school level question related to the "Mental arithmetic in the public school of S. Rachinsky" painting of N. B. Belsky: is [10, 11, 12, 13, and 14] the only series of five consecutive numbers, the sum of the squares of the first three of which is equal to the sum of the squares of the last two? | |
| Jan 10, 2014 at 21:04 | comment | added | KConrad | @ZurabSilagadze: строго говоря его зовут Сергей, не Семён. | |
| Jan 10, 2014 at 21:00 | history | edited | KConrad | CC BY-SA 3.0 |
added 23 characters in body
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| Jan 10, 2014 at 20:52 | comment | added | KConrad | I used this painting when teaching my number theory class four years ago. The task I proposed was not to compute the ratio, but rather to show the ratio on the board is an integer without having to calculate it explicitly. Since $365=5\cdot 73$, we want to show the numerator is a multiple of 5 and 73. Working mod 5 the numerator is 0 + 1 + 4 + 4 + 1, which is 0. Working mod 73, the numerator is 100 + 121 + 144 + 169 + 196 (I don't see a trick to find the squares mod 73 without their exact computation first), which is congruent to (27 + 48 -2) + (23 + 50) = 73 + 73, hence it's divisible by 73. | |
| Jan 4, 2014 at 21:42 | comment | added | Georges Elencwajg | What a wonderful combination of art and mathematics you are offering us, Zurab: thanks and +1. | |
| Jan 4, 2014 at 19:11 | vote | accept | Zurab Silagadze | ||
| Jan 4, 2014 at 17:49 | answer | added | abx | timeline score: 37 | |
| Jan 4, 2014 at 16:41 | comment | added | Zurab Silagadze | Another parametrization using Pythagorean triple (a,b,c) is (a,b,c,a+b,a-b) -- if you double the sum of two squares you get the sum of two squares (Lewis Carroll). For every $d$ and $e$ such that $d^2+e^2\ne 4^n(8m+7)$ for some integers $n$ and $m$, the Legendre's three-square theorem ensures the existence of some (a,b,c,d,e) Rachynsky quintet. | |
| Jan 4, 2014 at 15:52 | comment | added | Gerry Myerson | Maybe it's in Mordell's Diophantine Equations --- I'm currently without access, so I can't check. | |
| Jan 4, 2014 at 14:31 | comment | added | joro | If (a,b,e) is Pythagorean triple c=d completes the parametrization. | |
| Jan 4, 2014 at 14:22 | comment | added | joro | Don't know about complete parametrization, though partial parametrizations exist: a,b,c,d,e=(20, 9*x^2 + 9, 12*x^2 + 12, 9*x^2 + 25, 12*x^2) | |
| Jan 4, 2014 at 12:54 | history | asked | Zurab Silagadze | CC BY-SA 3.0 |