# 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.

## 3 Answers

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 '19 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.

There do not exist $$4$$ rational squares in arithmetic progression:

Theorem 1. Suppose that $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta$$ are rational numbers such that: $$\beta^2-\alpha^2=\gamma^2-\beta^2=\delta^2-\gamma^2.$$ Then $$\pm\alpha=\pm\beta=\pm\gamma=\pm\delta$$.

1.2. Elementary proof. The following elementary proof of theorem 1 is outlined by Dickson in , p. 440. As an exercise, fill in the details of the following steps.

I. To prove theorem 1 assume that we are given rational numbers $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta$$ satisfying the hypothesis of the theorem. We may assume that they are relatively prime, non-negative integers. The theorem then states that $$\alpha=\beta=\gamma=\delta$$. Assume that this is not true. Then we see that $$\alpha^2<\beta^2<\gamma^2<\delta^2$$.

Show that $$\gcd(\alpha,\beta)=\gcd(\beta,\gamma)=\gcd(\gamma,\delta)=1$$ and that $$\alpha$$, $$\beta$$, $$\gamma$$, $$\delta$$ are all odd. Conclude that $$\gcd(\beta+\alpha,\beta-\alpha)=\gcd(\gamma+\beta,\gamma-\beta)=\gcd(\delta+\gamma,\delta-\gamma)=2$$.

II. Defining positive integers $$a,b,c,d$$ by: $$2a:=\gcd(\beta+\alpha,\gamma+\beta),\\ 2b:=\gcd(\beta+\alpha,\gamma-\beta),\\ 2c:=\gcd(\beta-\alpha,\gamma+\beta),\\ 2d:=\gcd(\beta-\alpha,\gamma-\beta),$$ show that we have: $$\beta+\alpha=2ab,\\ \beta-\alpha=2cd,\\ \gamma+\beta=2ac,\\ \gamma-\beta=2bd,\\ \delta+\gamma=2bc,\\ \delta-\gamma=2ad.$$

III. Then $$(a+d)b=(a-d)c$$ and $$(c+d)a=(c-d)b$$, and we have the number $$\gcd(a+d,a-d)$$ and $$\gcd(c+d,c-d)$$ are both either $$1$$ or $$2$$. We can then conclude that $$a+d=mc,\\a-d=mb,\\c+d=nb,\\c-d=na,$$ with $$m,n\in\{1,2\}$$.

IV. The conclusion in the last step is incompatible with the positivity of $$a,b,c,d$$.

• 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 '19 at 12:06
• @SasQ, see my edited answer. – lhf Jul 10 '19 at 12:21

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