The Problem is that:
Show that there is a positive integer $ n $ such that $$ \left|\left\{\{x, y, z\}: x, y, z \in \mathbb{Z}^{+}, x<y<z, x^{2}+y^{2}+z^{2}=n\right\}\right| \geq 2021 $$
I tried to solve it by consider the 3-subset $\in \{1, 2, ..., \sqrt{n}\}$, but calculate the numbers of $x^{2}+y^{2}+z^{2}=n$ seems hard to me. Any hints or solutions are appreciated!
Consider an $N\gg1$, and let $B$ be the ball of radius $\sqrt{N}$ around the origin. The part $B_+$ of this ball containing the points with $0<x<y<z$ has volume $${\rm vol}(B_+)={\pi N^{3/2}\over 36}\ .$$ The number of lattice points $(x,y,z)$ lying in $B_+$ is about equal to this volume (I'm omitting the approximation estimates). All of these lattice points have integer $x^2+y^2+z^2\in[N]$. By the pigeon principle there is then an $n\in[N]$ such that at least $${{\rm vol}(B_+)\over N}={\pi\sqrt{N}\over36}$$ of them have $x^2+y^2+z^2=n$. Choosing $$N\geq\left({36\cdot2021\over\pi}\right)^2$$ then proves the claim.