5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $5 x^{2}-6 x y+y^{2}=6^{100}$.
Answer: 19594.

Solution: By factoring the left and right sides of the equation, we get $(5 x-y)(x-y)=2^{100} \cdot 3^{100}$. Since each factor on the left side is an integer, it follows that

$$
\left\{\begin{array}{l}
5 x - y = 2 ^ { k } \cdot 3 ^ { l }, \\
x - y = 2 ^ { 1 0 0 - k } \cdot 3 ^ { 1 0 0 - l }
\end{array} \text { or } \left\{\begin{array}{l}
5 x-y=-2^{k} \cdot 3^{l}, \\
x-y=-2^{100-k} \cdot 3^{100-l}
\end{array}\right.\right.
$$

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

$$
\left\{\begin{array}{l}
x=2^{k-2} \cdot 3^{l}-2^{98-k} \cdot 3^{100-l} \\
y=2^{k-2} \cdot 3^{l}-5 \cdot 2^{98-k} \cdot 3^{100-l}
\end{array}\right.
$$

Consider the first equation. The exponents in the powers of three are non-negative. The sum of the exponents in the powers of two is 96, 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 $2 \leq k \leq 98,0 \leq l \leq 100-$ a total of $97 \cdot 101=9797$ options.

The second system also has 9797 solutions; thus, there are 19594 solutions in total.