Proving that $n^2-l$, $n^2$ and $n^2+l$ can't all be perfect squares

Question

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.

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. — Andrés E. Caicedo, Oct 09 '13 at 22:15

score 32 · Accepted Answer · answered Oct 09 '13 at 21:02

32

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

answered Oct 09 '13 at 21:02

Arthur

5,524

score 10 · Answer 2 · answered Oct 09 '13 at 22:11

10

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.

answered Oct 09 '13 at 22:11

Kieren MacMillan

7,889
2
28
70

Proving that $n^2-l$, $n^2$ and $n^2+l$ can't all be perfect squares

2 Answers2