8. (40 points) At the "Horizon" base, 175 university students arrived. Some of them are acquainted with each other, while others are not. It is known that any six students can be accommodated in two three-person rooms such that all those in the same room are acquainted with each other. What is the minimum number of pairs of acquainted students that could be among those who arrived at the base?
Answer: 15050.

Solution: It is clear that each student has no more than three unfamiliar students. If each student has no more than two unfamiliar students, then each is familiar with at least 172 students, and the total number of pairs of familiar students is no less than $175 \times 172 / 2 = 15050$. Suppose there is a student $A$ who has three unfamiliar students $B_{1}, B_{2}$, and $B_{3}$. Consider the group of six students: $A, B_{1}$, $B_{2}$, $B_{3}$, and two other arbitrary students $C$ and $D$. They can be seated in two three-person rooms, so $A$ must be in the same room with $C$ and $D$, which means $C$ and $D$ are familiar with each other and with $A$. Therefore, any two students, different from $B_{1}$, $B_{2}$, and $B_{3}$, are familiar with each other. In particular, if some student, different from $B_{1}$, $B_{2}$, and $B_{3}$, is unfamiliar with someone, then this someone is one of $B_{1}$, $B_{2}$, and $B_{3}$. But each $B_{i}$ is unfamiliar with no more than three students, so the number of pairs of unfamiliar students is no more than nine. Therefore, in this case, the number of pairs of familiar students is no less than $175 \cdot 174 / 2 - 9 = 15216 > 15050$. We will now show that 15050 pairs of students can exist. Seat all students at a large round table and make everyone familiar except those sitting next to each other. It is not difficult to verify that this construction satisfies the condition of the problem.