# Probability distribution from standard domain (multiple pairs single prime) - V

Pick a random pair $$(a,b)\in\mathbb Z_n^2\backslash\{0,0\}$$. Denote $$N_2(a,b,n)$$ to be minimum $$\ell_2$$ norm of vector $$(x,y)$$ as $$(x,y)$$ ranges over all non-zero integral solutions to $$(x,y)\equiv t(a,b)\bmod n$$ where $$t\in\mathbb Z$$ with $$0.

This is the result from Akshay Venkatesh when $$n$$ is prime. Then it is true as $$n\rightarrow\infty$$ the distribution of $$N_2(a,b,n)/\sqrt{n}$$ coincides with distribution of $$1/\sqrt y$$ where $$x+iy$$ is picked at random with respect to hyperbolic measure from $$\{z:|z|\geq1,|\mathcal R(z)|\leq\frac12\}$$.

Hence It looks like given $$\epsilon>0$$ and prime $$n$$ large enough and randomly taken there should be coprime $$a,b$$ with $$\sqrt n with $$N_2(a,b,n)/\sqrt n.

Pick $$k=O(1)$$ random pairs $$(a_1,b_1),(a_2,b_2),\dots,(a_k,b_k)\in\mathbb Z_n^2\backslash\{0,0\}$$ where $$k\geq3$$. What is the probability that $$\frac{\max(N_2(a_1,b_1,n),N_2(a_2,b_2,n),\dots,N_2(a_k,b_k,n))}{\sqrt n}<\frac{n^\epsilon}{\sqrt n}?$$

What I want validation is on following possibility:

1. Can we get $$k=O(1)$$ with $$k\geq3$$ such coprime pairs of $$(a_1,b_1),(a_2,b_2),\dots,(a_k,b_k)$$ for all primes $$n$$ large enough with $$\sqrt n and with $$\frac{\max(N_2(a_1,b_1,n),N_2(a_2,b_2,n),\dots,N_2(a_k,b_k,n))}{\sqrt n}<\frac{n^\epsilon}{\sqrt n}?$$

2. Slightly different version is can we get $$2k=O(1)$$ with $$k\geq3$$ such mutually coprime integers $$a_1,b_1,a_2,b_2,\dots,a_k,b_k$$ for all primes $$n$$ large enough with $$\sqrt n and with $$\frac{N_2(a_1,b_1,a_2,b_2,\dots,a_k,b_k,n)}{\sqrt n}<\frac{n^\epsilon}{\sqrt n}$$ where $$N_2(a_1,b_1,a_2,b_2,\dots,a_k,b_k,n)$$ is direct generalization of $$N_2(a,b,n)$$.

There is a single $$t$$ in 2. while three different $$t$$'s are allowed in 1.

I only need the result at $$k=3$$.