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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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3 Answers 3

up vote 14 down vote accepted

Here are a few proofs: 1, 2 (which is excellent), and the somewhat bizarre 3.

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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2  
Oh, this is excellent! Many thanks! –  Bruce George Jun 6 '11 at 2:03
1  
+1 for the bizarre link. –  Arjang Jun 6 '11 at 7:28

A quick Google search found this: http://www.math.ku.dk/~kiming/lecture_notes/2007-2008-elliptic_curves/4_squares_in_arithmetic_progression.pdf . It contains a sketch of an elementary proof at the end and cites Dickson's History of the theory of numbers.

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Very nice write-up. Thank you! –  Bruce George Jun 6 '11 at 2:04

My favourite proof of this is Van der Poorten's (http://maths.mq.edu.au/~alf/SomeRecentPapers/183.pdf) — it uses descent, as Fermat almost certainly would have.

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