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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.

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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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    @KCd, the link is still available through the Internet Archive: [link](http://web.archive.org/web/20130604030207/http://www.math.niu.edu/~rusin/known-math/96/fermatlike).2016-02-25
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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.

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    What a terrific idea! I did, and the author says the link should be restored soon. He also says his treatment is the same as in Cassels' book on elliptic curves.2014-10-21
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My favourite proof of this is Van der Poorten's — it uses descent, as Fermat almost certainly would have.

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    @GerryMyerson: Thanks! Updated.2017-08-29