I am not sure if this problem is of the appropriate difficulty for math overflow, but here it is.


Suppose we are considering pairs $(x,y)$ with $1 \leq x,y \leq p-1$ for some prime $p$. As points over $\mathbb{F}\_p^2$, we can define the usual projective equivalence relation where we consider $(x,y)$ and $(x', y')$ to be equivalent if there exists non-zero $\rho \in \mathbb{F}\_p$ such that $(x,y) \equiv (\rho x', \rho y') \pmod{p}$. It is easy to show that every point $(x,y)$ with $1 \leq x,y \leq p-1$ is equivalent to $(1, z)$ for some $z$. 


My question is, for a given $p$, how many pairs $(x,y)$ (as above) are equivalent to a point of the form $(u,v)$, where $1 \leq u, v < \sqrt{p}$? It is not true that all pairs work for every $p$. For example, if $p = 11$ and $(x,y) = (3,5)$, then $3\rho \in [1, 3]$ for $\rho = 1, 4, 8$. However, $5\rho \in [1,3]$ for $\rho = 5,7,9$, and these two sets have empty intersection. Hence $(3,5)$ is not equivalent to a point of the form $(u,v)$ with $1 \leq u,v \leq 3$.


Can anything useful be said? I am interested to know, ideally, conditions to ensure that $(x,y)$ is equivalent to a point in the box $[1, \sqrt{p}) \times [1, \sqrt{p})$ and if this is not available, an estimate for the size of the set of exceptions.


Thanks for any insights.



A question regarding simultaneous congruences
The size of the exceptional set is $\Omega(p^2)$ and indeed, there are at most $\frac6{\pi^2}(1+o(1))p^2$ pairs, projectively equivalent to a point inside the square $Q:=[1,N]\times[1,N]$, where $N=\lfloor\sqrt p\rfloor$. To see this, notice that if a point from $Q$ is an integer multiple of another point from $Q$, then the two points determine the same equivalence class. Hence, the number of classes, determined by the points of $Q$, does not exceed the number of points of $Q$ with co-prime coordinates, and this number is known to be $\frac6{\pi^2}(1+o(1))N^2$ as $N\to\infty$. Finally, every class contains exactly $p-1$ points $(x,y)\in{\mathbb F}\_p$, totalling to at most $\frac6{\pi^2}(1+o(1))N^2(p-1)=\frac6{\pi^2}(1+o(1))p^2$ points, equivalent to a point from $Q$.


