# Problem 4.

In the Oxy plane, find the smallest and largest distances between two points $(x ; y)$, the coordinates of which are integers and satisfy the equation $y^{2}=4 x^{2}-15$.


Solution. Rewrite the given equation as $(2 x-y)(2 x+y)=1 \cdot 3 \cdot 5$. We see that the integer points satisfying this equation and lying in the first quadrant are solutions to the following systems:

$$
\left\{\begin{array} { l } 
{ 2 x - y = 1 , } \\
{ 2 x + y = 1 5 }
\end{array} \text { or } \left\{\begin{array}{l}
2 x-y=3 \\
2 x+y=5
\end{array}\right.\right.
$$

These are the points $(4 ; 7)$ and $(2 ; 1)$. The other points lying in the $O x y$ plane are symmetrically located relative to the coordinate axes and the origin.

As the nearest points, we can take the pair $(2 ; 1)$ and $(2 ;-1)$, the distance between them is 2. The most distant pair is $(4 ; 7)$ and $(-4 ;-7)$, the distance between which is $2 \sqrt{4^{2}+7^{2}}=2 \sqrt{65}$.

Answer: $2 ; 2 \sqrt{65}$.