Counting in their heads - a painting of Bogdanov-Belsky http://s50.radikal.ru/i127/1112/fc/c8b4f15ff154.jpg

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?

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?

Counting in their heads - a painting of Bogdanov-Belsky http://s50.radikal.ru/i127/1112/fc/c8b4f15ff154.jpg

This 1895 painting of Nikolai Bogdanov-Belsky shows mental calculations in the public school of Semyon 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?

Counting in their heads - a painting of Bogdanov-Belsky http://s50.radikal.ru/i127/1112/fc/c8b4f15ff154.jpg

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?

Counting in their heads - a painting of Bogdanov-Belsky http://s50.radikal.ru/i127/1112/fc/c8b4f15ff154.jpg

This 1895 painting of Nikolai Bogdanov-Belsky describes mental arithmetics in the public school of Semyon 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 equality $10^2+11^2+12^2=13^2+14^2$. Now this Rachynsky 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 Rachynsky 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), $a^2+b^2=c^2$, 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 $m-n$ is odd. Can one establish an analogues result for Rachynsky quintets?

Counting in their heads - a painting of Bogdanov-Belsky http://s50.radikal.ru/i127/1112/fc/c8b4f15ff154.jpg

This 1895 painting of Nikolai Bogdanov-Belsky shows mental calculations in the public school of Semyon 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?