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.

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    $\begingroup$ Oh, this is excellent! Many thanks! $\endgroup$ – Bruce George Jun 6 '11 at 2:03
  • $\begingroup$ 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. $\endgroup$ – KCd Jun 7 '15 at 21:34
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    $\begingroup$ @KCd, the link is still available through the Internet Archive: link. $\endgroup$ – FredH Feb 25 '16 at 16:40
  • $\begingroup$ Link number 2 is also gone. No backup in Internet Archive. $\endgroup$ – 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 [9], 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 $a,b,c,d$.

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


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