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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?

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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.

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  • $\begingroup$ You could accept this (you may have to wait a bit) to show it answers your question. $\endgroup$ – Ross Millikan Nov 21 '13 at 15:16
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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 $a=120$.

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  • $\begingroup$ Sorry but I cannot see how your answer solves the question. $\endgroup$ – David Cushing Nov 20 '13 at 17:20
  • $\begingroup$ 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. $\endgroup$ – PITTALUGA Nov 20 '13 at 17:28
  • $\begingroup$ But a is not fixed. I am asking if there exists infinitely many x, d coprime such that x,x+d,x+2d,x+4d are all square numbers or if 49,169,289,529 is the only such sequence $\endgroup$ – David Cushing Nov 20 '13 at 17:45

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