4 squares almost in an arithmetic progression

Question

It is well known that the exists no arithmetic progression of squares of length $4$. But consider the following arithmetic progression of length $5$:

$49,169,289,409,529$.

All terms apart from the $4^{th}$ term are squares. Does there exist infinitely many of these progressions that are not just a multiple of the one above? Or are they even anymore apart from the listed one?

Answer 1

I have since posted this question to math overflow (https://mathoverflow.net/questions/149527/4-squares-almost-in-an-arithmetic-progression) and a mathematician has solved it.

Answer 2

It cannot exist. In fact let $$ a=n^2-m^2=(n-m)(n+m) $$ and you have finite possibilities for $a$ to be generated as a multiplication of a difference times a sum.

In your example, you chose $a=120$.