# Rachinsky quintets

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?

• What a wonderful combination of art and mathematics you are offering us, Zurab: thanks and +1. Jan 4 '14 at 21:42
• 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 10 '14 at 20:52
• @ZurabSilagadze: строго говоря его зовут Сергей, не Семён. Jan 10 '14 at 21:04
• 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)
– user111753
Jun 30 '17 at 11:10
• 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.
– user111753
Jun 30 '17 at 11:10

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.

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).