4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $x^{2}+x y=30000000$.
Answer: 256.

Solution. By factoring the left and right sides of the equation, we get $x(x+y)=$ $3 \cdot 2^{7} \cdot 5^{7}$. Then, if $x>0$, $x$ is one of the divisors of the right side. The right side has a total of $2 \cdot 8 \cdot 8=128$ divisors (since any divisor can be represented as $3^{a} \cdot 2^{b} \cdot 5^{c}$, where $a, b$, and $c$ are non-negative integers not exceeding 1, 7, and 7, respectively, i.e., there are 2 ways to choose $a$, 8 ways to choose $b$, and 8 ways to choose $c$). Note that if the right side is divisible by $x$, then it automatically follows that $y \in \mathbb{Z}$, and $y$ is uniquely determined. Therefore, there are a total of $2 \cdot 128=256$ pairs of numbers.