8. Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be distinct positive integers less than 2021, then the number of arrays $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)$ that satisfy
$$
\left(a_{1}{ }^{2}+a_{2}{ }^{2}+a_{3}{ }^{2}+a_{4}{ }^{2}\right)\left(a_{2}{ }^{2}+a_{3}{ }^{2}+a_{4}{ }^{2}+a_{5}{ }^{2}\right)=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{5}\right)^{2}
$$

is $\qquad$
Answer: 424
Difficulty: Hard
Examination: Cauchy Inequality, Geometric Sequence, Combinatorial Counting
Analysis: By Cauchy Inequality, the necessary and sufficient condition for the equality to hold is $\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{3}}{a_{4}}=\frac{a_{4}}{a_{5}}$, i.e., $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ form a geometric sequence. Since they are distinct, the common ratio $q \neq 1$. By symmetry and the common ratio being a rational number, we can assume $q=\frac{s}{t},(s, t)=1, s>t \geq 1$. At this point, $a_{1}t \geq 1$, so the number of arrays that satisfy the condition is
$$
2 \times[1 \times(1+\cdots+5)+2 \times(1+\cdots+4)+4 \times(1+2+3)+17 \times(1+2)+102 \times 1]=424 \text {  }
$$