Marinada '25

[lang=hr] Nađi prirodne brojeve $x,y,z$ tako da $x>z>1999\cdot2000\cdot2001>y$ i vrijedi jednadžba: $$ 2000x^2+y^2=2001z^2 $$ Kao rješenje upiši trojku $(x,y,z)$ (bez zagrada, odvojeno zarezima) gdje je $y$ minimalan moguć i zbroj $x+z$ je minimalan moguć za taj najmanji $y$. [/lang] [lang=en] Find natural numbers $x,y,z$ such that $x>z>1999\cdot2000\cdot2001>y$ and such that they satisfy the following equation: $$ 2000x^2+y^2=2001z^2 $$ For the solution write the triple $(x,y,z)$ (without the brackets, separated by commas) where $y$ is as small as possible and $x+z$ is also as small possible for that minimal $y$. [/lang]