TITLE: Proving that $n^2-l$, $n^2$ and $n^2+l$ can't all be perfect squares
QUESTION [9 upvotes]: 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.

REPLY [10 votes]: 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.