Let f(n) be the maximum number of squares in an AP (arithmetic progression) of length n. For example, $f(3)=3$, as 1, 25, 49 is a 3-term arithmetic progression with three squares, and $f(4)=3$, as there are no 4 term arithmetic progressions of squares. Also, $f(5)=4$, with the AP 49, 169, 289, 409, 529 as a small example.
Trivially, f is monotone increasing, as adding terms onto an existing AP cannot reduce the number of squares. Also, $f(a+b) \leq f(a)+f(b)$, by concatenation of sequences. It seems to me that the easiest way to find upper bounds on f is to constrain configurations of squares. Let $(0, a, b, c)$ (with $0<a<b<c$ denote a configuration of squares of the form: $M, M+aK, M+bk, M+ck$, where $k>0$. The configuration $(0, 1, 2, 3)$ is a four term arithmetic progression, which we already know is ruled out. Using elliptic curves you can show that $(0, 1, 3, 4)$ and $(0, 1, 4, 5)$ are also impossible (and it looks like many more are impossible as well. On the positive end, there are solutions for any configuration $(0, a, b, c)$ when $c \neq a+b$ (I'm working on a parametric solution).
By eliminating those configurations, I have found that $f(6)=4$, $4 \leq f(7) \leq 5$, and $f(8)=5$ with $1, 25, 49, 73, 97, 121, 145, 169$ as an example.
Up to what $n$ is $f(n)$ known? Specifically, are f(9) and f(10) known?