4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $y^{2}-x y=700000000$.
Answer: 324.

Solution. Factoring the left and right sides of the equation, we get $y(y-x)=$ $7 \cdot 2^{8} \cdot 5^{8}$. Then if $y>0$, $y$ is one of the divisors of the right side. The right side has a total of $2 \cdot 9 \cdot 9=162$ divisors (since any divisor can be represented as $7^{a} \cdot 2^{b} \cdot 5^{c}$, where $a, b$, and $c$ are non-negative integers not exceeding 1, 8, and 8, respectively, i.e., there are 2 ways to choose $a$, 9 ways to choose $b$, and 9 ways to choose $c$). Note that if the right side is divisible by $y$, then it automatically follows that $x \in \mathbb{Z}$, and in this case, $x$ is uniquely determined. Therefore, there are a total of $2 \cdot 162=324$ pairs of numbers.