Let $n$ be an integer greater than or equal to 3. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.
We solve this problem by employing the construction method and claiming the following $n$ points met the requirements: $$(1, 1), (2, 4), (3, 9), \cdots, (n, n^2)$$
Any two points in this set can be represented by $(a, a^2)$ and $(b, b^2)$ where $1\le a < b \le n$. The, their distance: $$\sqrt{(a-b)^2+(a^2-b^2)^2}=(a-b)\sqrt{1+(a+b)^2}$$
This cannot be rational because $1+(a+b)^2$ cannot be a sqaure.
Meanwhile, the area of any triangle whose vertex are $(a, a^2)$, $(b, b^2)$, and $(c, c^2)$ where $1\le a < b < c \le n$ can be calculated as $$\frac{1}{2}|ab^2+bc^2+ca^2-a^2b-b^2c-c^2a|$$ which clearly is rational because $a$, $b$ and $c$ are all integers.