# Squares in arithmetic progression

It is easy to find 3 squares (of integers) in arithmetic progression. For example, $1^2,5^2,7^2$.

I've been told Fermat proved that there are no progressions of length 4 in the squares. Do you know of a proof of this result?

(Additionally, are there similar results for cubes, 4th powers, etc? If so, what would be a good reference for this type of material?)

Edit, March 30, 2012: The following question in MO is related and may be useful to people interested in the question I posted here.

Here are a few proofs: 1, and the somewhat bizarre 3. I'd previously linked to Kiming's exposition to prove this result, but the link has been removed. This is the proof described in lhf's answer --- and I think of this as a very elementary approach.

Unfortunately, there are no cases where you have nontrivial arithmetic progressions of higher powers. This is a string of proofs. Carmichael himself covered this for n = 3 and 4, about a hundred years ago. But it wasn't completed until Ribet wrote a paper on it in the 90s. His paper can be found here. The statement is equivalent to when we let $$\alpha = 1$$. Funny enough, he happens to have sent out a notice on scimath with a little humor, which can still be found here.

• Oh, this is excellent! Many thanks! – Bruce George Jun 6 '11 at 2:03
• The link at the end from sci.math is now broken. It looks like the whole "Mathematical Atlas" site that Rusin set up and hosted on his webpages is now gone. – KCd Jun 7 '15 at 21:34
• @KCd, the link is still available through the Internet Archive: link. – FredH Feb 25 '16 at 16:40
• Link number 2 is also gone. No backup in Internet Archive. – SasQ Jul 10 at 11:56

A quick Google search found this paper: On 4 Squares in Arithmetic Progression by Ian Kiming. It contains a sketch of an elementary proof at the end and cites Dickson's History of the theory of numbers. It is reproduced below.

• Very nice write-up. Thank you! – Bruce George Jun 6 '11 at 2:04
• The link seems to be dead. Does anyone have a backup or another link to the exposition? – PrimeRibeyeDeal Oct 19 '14 at 16:42
• @PrimeRibeyeDeal, that's too bad. Perhaps you can ask the author Ian Kiming ? – lhf Oct 19 '14 at 19:03
• @SasQ, here is a copy: pdfs.semanticscholar.org/9fcc/… – lhf Jul 10 at 12:06
• @SasQ, see my edited answer. – lhf Jul 10 at 12:21

My favourite proof of this is Van der Poorten's — it uses descent, as Fermat almost certainly would have.