# 4 squares almost in an arithmetic progression

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?

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 $a=120$.
• If I understood correctly your problem, you want squares in arithmetic progression, right? In your example, take $a=120=13^2-7^2=17^2-13^2=...$, but you cannot continue forever since $a=(p-q)(p+q)$ (in your case you chose primes...) so for a FIXED $a$ you have a finite number of possibilities. – PITTALUGA Nov 20 '13 at 17:28