Probability that a random pair of points are opposite corners of a square in an $n\times n$ integer lattice

For given $k$, $1\leq k\leq n-1$, an axis-aligned square of side length $n-k$ can be placed in $k^2$ ways, and each such square hosts $n-k$ squares with vertices on its rim. The total number $Q_n$ of admissible squares therefore is given by $$Q_n=\sum_{k=1}^{n-1}k^2(n-k)=\ldots={n^2(n^2-1)\over12}\ .$$ I think the exact form of the final result is a coincidence, cf. the formula for $\sum_{k=1}^n k^3$. – Working with the individual choices is made more difficult by the following fact: The chosen points have to fulfill a parity condition before you even can think of a possible square. If $n$ is odd the probability that this parity condition holds is $\ne{1\over2}$.

By the way: Say "a random pair of lattice points" since two points chosen independently can coincide.