# Probability density from standard domain (Typical Box principle and Chinese Remainder Theorem) - III

Pick a random pair $$(a,b)\in\mathbb Z_n^2\backslash\{0,0\}$$. Denote $$N_2(a,b,n)$$ to be minimum $$\ell_r$$ 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.

Now let integers $$a,b$$ be of size in $$(\sqrt m,m-\sqrt m)$$.

Let $$p_1,p_2$$ be primes of similar size some integer $$m$$ and let $$t_1$$ attain $$N_2(a,b,p_1)$$ and $$t_2$$ attain $$N_2(a,b,p_2)$$. From Dirichlet pigeonhole we know $$N_2(a,b,p_1)$$ and $$N_2(a,b,p_2)$$ are less than $$\sqrt{2m}$$.

Let $$t$$ attain $$N_2(a,b,p_1p_2)$$ (note $$(t,p_1p_2)=1$$ need not hold while $$(t_1,p_1)=(t_2,p_2)=1$$ holds).

Then $$N_2(a,b,p_1p_2)$$ seems to be typically $$m\sqrt 2$$.

1. Is it possible for $$N_2(a,b,p_1p_2)$$ to be smaller than $$\sqrt{2m}-\mu$$ if $$N_2(a,b,p_1)$$ and $$N_2(a,b,p_2)$$ are greater than $$\sqrt{2m}-\mu$$ at a $$\mu\in(0,\sqrt{2m})$$?

2. If so what is the probability?

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 $$\{z:|z|\geq1,|\mathcal R(z)|\leq\frac12\}$$.