Given the positive integers, $x$, $y$, and $z$, are consecutive terms of an arithmetic progression, the least value of the positive integer, $n$, for which the equation, $x^2-y^2-z^2=n$, has exactly two solutions is $n=27$: $${34}^2-{27}^2-{20}^2={12}^2-9^2-6^2=27.$$

It turns out that $n=1155$ is the least value which has exactly ten solutions.

How many values of $n$ less than one million have exactly ten distinct solutions?