Mongolian Mathematical Olympiad

Note that $(3n{)}^2+(4n{)}^2=(5n^2)$ from where we derive $(3n{)}^2+(4n-1{)}^2-(5n-1{)}^2=2n$ and $(3n+2{)}^2+(4n{)}^2-(5n-1{)}^2=2n+3$.

a. Let's consider the case of quadratical number is odd. Set $a=3n+2$, $b=4n$, $c=5n+1$ in the equality $(3n+2{)}^2+(4n{)}^2-(5n-1{)}^2=2n+3$. If $n\ge 3$ then the condition $a<b<c$ holds. Therefore if $m\ge 9$ then any odd number $m=2n+3$ can be written in the form $m=2n+3=a^2+b^2+c^2$ as noted above. Now consider the cases $m=1,3,5,7$.

$1=4^2+7^2-8^2$, $3=4^2+6^2-7^2$, $5=4^2+5^2-6^2$, $7=10^2+14^2-17^2$. Thus every odd number is quadratical.

b. Let's consider the case of quadratical number is even. Set $a=3n$, $b=4n-1$, $c=5n-1$ in the equality $(3n{)}^2+(4n-1{)}^2-(5n-1{)}^2=2n$. If $n\ge 2$ then the condition $a<b<c$ holds. Note that $2=5^2+11^2-12^2$ and from where we conclude that every even number is quadratical.

Hence all numbers 1 - 2014 are quadratical.