Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 9 '13 at 22:15
up vote 31 down vote accepted

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

share|cite|improve this answer

Indeed, Fermat proved that there are infinite triples of squares in arithmetic progression, and gave formulas (parameterizations) to generate them — see for example

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.