5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $6 x^{2}-7 x y+y^{2}=10^{100}$.
Answer: 19998.

Solution: By factoring the left and right sides of the equation, we get $(6 x-y)(x-y)=2^{100} \cdot 5^{100}$. Since each factor on the left side is an integer, it follows that

$$
\left\{\begin{array}{l}
6 x - y = 2 ^ { k } \cdot 5 ^ { l } , \\
x - y = 2 ^ { 1 0 0 - k } \cdot 5 ^ { 1 0 0 - l }
\end{array} \text { or } \left\{\begin{array}{l}
6 x-y=-2^{k} \cdot 5^{l} \\
x-y=-2^{100-k} \cdot 5^{100-l}
\end{array}\right.\right.
$$

where $k$ and $l$ are integers from the interval $[0 ; 100]$.

Let's find the number of solutions to the first system. Expressing $x$ and $y$ from it, we get

$$
\left\{\begin{array}{l}
x=2^{k} \cdot 5^{l-1}-2^{100-k} \cdot 5^{99-l} \\
y=2^{k} \cdot 5^{l-1}-6 \cdot 2^{100-k} \cdot 5^{99-l}
\end{array}\right.
$$

Consider the first equation. The exponents of the powers of two are non-negative. The sum of the exponents of the powers of five is 98, so at least one of them is positive, i.e., the corresponding term is an integer. Since the left side of the equation is also an integer, the second term on the right side of the equation must also be an integer. Therefore, for the existence of integer solutions, it is necessary and sufficient that $0 \leq k \leq 100,1 \leq l \leq 99$ - a total of $99 \cdot 101=9999$ options.

The second system also has 9999 solutions; thus, there are 19998 solutions in total.