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Counting in their heads - a painting of Bogdanov-Belsky

This 1895 painting of Nikolai Bogdanov-Belsky shows mental calculations in the public school of Sergei Rachinsky. Boys in a Russian village school try to calculate $(10^2+11^2+12^2+13^2+14^2)/365$ in their heads. One of the methods of solution is based on the equality $10^2+11^2+12^2=13^2+14^2$. Now this Rachinsky equality can be considered as a generalization of the well-known Pythagorean triple (3,4,5), $3^2+4^2=5^2$, and in analogy with the Pythagorean triples one can define Rachinsky quintets as a set of five positive integers $(a,b,c,d,e)$ such that $a^2+b^2+c^2=d^2+e^2$. It is known that all primitive Pythagorean triples $(a,b,c)$ such that $a^2+b^2=c^2$ are generated by Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$, where $m$ and $n$ are positive integers such that $m>n$, $m$ and $n$ are coprime, and $m \not\equiv n \bmod 2$. Can one establish an analogous result for Rachinsky quintets?

Zurab Silagadze
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  • Don't know about complete parametrization, though partial parametrizations exist: a,b,c,d,e=(20, 9x^2 + 9, 12x^2 + 12, 9x^2 + 25, 12x^2) – joro Jan 04 '14 at 14:22
  • If (a,b,e) is Pythagorean triple c=d completes the parametrization. – joro Jan 04 '14 at 14:31
  • Maybe it's in Mordell's Diophantine Equations --- I'm currently without access, so I can't check. – Gerry Myerson Jan 04 '14 at 15:52
  • 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. – Zurab Silagadze Jan 04 '14 at 16:41
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    What a wonderful combination of art and mathematics you are offering us, Zurab: thanks and +1. – Georges Elencwajg Jan 04 '14 at 21:42
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    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. – KConrad Jan 10 '14 at 20:52
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    @ZurabSilagadze: строго говоря его зовут Сергей, не Семён. – KConrad Jan 10 '14 at 21:04
  • 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? – José Hdz. Stgo. Jan 12 '14 at 01:07
  • @KConrad: Спасибо! Википедия ввел в заблуждение: http://en.wikipedia.org/wiki/Nikolay_Bogdanov-Belsky – Zurab Silagadze Jan 17 '14 at 12:33
  • Whenever $\ a^2+b^2=c^2,\ $ we get: $$ a^2 + b^2 + c^2\ =\ (a-b)^2 + (a+b)^2 $$ Enjoy! – Wlod AA Jun 30 '17 at 12:37
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    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 30 '17 at 11:10
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    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. –  Jun 30 '17 at 11:10
  • My wife's grandmother used to tell me how she and her siblings made those shoes for themselves out of reeds. – Ben McKay Dec 11 '21 at 09:03

2 Answers2

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The equation $a^2+b^2+c^2=d^2+e^2$ defines a quadric $Q\subset\mathbb{P}^4$, with a rational point $p=(1,0,0,0,1)$. Therefore it is rational : projecting from $p$, say on the hyperplane $e=0$, defines a birational map $Q --> \mathbb{P}^3$. The inverse of that map, namely $$ (x,y,z,t)\mapsto (x-\lambda ,y,z,t,\lambda )\quad \mbox{with }\lambda :=(x^2+y^2+z^2-t^2)/2x$$ gives a parametrization of all rational points in $Q$ with $x\neq 0$; to get integral points just multiply all coordinates by $2x$. To get the remaining points replace $p$ by $p'=(0,1,0,0,1)$, etc.

abx
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    Nice answer! I found that the projective geometry behind it is explained in these lectures http://www.math.chalmers.se/~ulfp/Teaching/geometri.pdf – Zurab Silagadze Jan 04 '14 at 19:15
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The following recipe (algorithm) generates all solutions. It may be viewed as a parametrization in a general(ized) sense.

W.l.o.g. we may assume that $c$ is odd. Then

$$ \left(\frac{x-y}2\right)^2 + \left(\frac{u-v}2\right)^2 + c^2 \ =\ \left(\frac{x+y}2\right)^2 + \left(\frac{u+v}2\right)^2 $$

where three conditions hold:

  1. $\ x\equiv y\equiv 1\ \mbox{mod}\ 2$
  2. $\ u\equiv v\equiv 0\ \mbox{mod}\ 2$
  3. $\ u\cdot v = c^2-x\cdot y$

i.e. we may take arbitrary $x$ and $y$ as in condition 1, and then one decomposes $c^2-x\cdot y$ (see condition 3), where $\ u\ v\ $ are as in condition 2; of course $\ 4\,|\,c^2-x\cdot y\ $ (and the expressions under the squares are integers).

Wlod AA
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