I tried to write a proof and used the argument that if $n^2$ is a perfect square, $n^2-l$ and $n^2+l$ can't both be perfect squares. However, I can't find a proof for this statement. Can you help me with this?

What I have tried: Suppose that $n^2-l$, $n^2$ and $n^2+l$ are all perfect squares. Then this must hold $$n^2-l = \sum_{i=0}^{m-a}(2i+1)$$ $$n^2 = \sum_{i=0}^{m}(2i+1)$$ $$n^2+l = \sum_{i=0}^{m+b}(2i+1)$$

From first two I can obtain that $$l=\sum_{i=m-a+1}^{m}(2i+1)$$ And from last two: $$l=\sum_{i=m+1}^{m+b}(2i+1)$$ While it seems that these both can't hold, I am not able to show an obvious contradiction there.

I also got a suggestion to start with this: $$n^2+l=x^2$$ $$n^2-l=y^2$$ I tried to count them together but $2n^2=x^2+y^2$ also doesn't seem a straightforward contradiction to me.

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    $\begingroup$ As mentioned in the answers, the result you mention is not true. What is true is that no $4$ squares can be in arithmetic progression, see here. $\endgroup$ – Andrés E. Caicedo Oct 9 '13 at 22:15

What you're trying to prove is false. A counterexample is given by $n=5$ and $l = 24$.


Indeed, Fermat proved that there are infinite triples of squares in arithmetic progression, and gave formulas (parameterizations) to generate them — see for example http://cms.math.ca/crux/v23/n5/page274-277.pdf.


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