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 exceptions.

Thanks for any insights.

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.