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Recently reading some papers motivated by Fermat's more well-known result that no four squares can be in an arithmetic progression (with nonzero common difference!), I saw the claim repeated multiple times that Legendre proved here that no three cubes can be arithmetic progression (for example, this is claimed at the beginning of this paper).

Having tried to scour through that enormous 490 page document which, by modern standards, is rather difficult to navigate and using my reasonable French I have not yet found the proof.

Can someone either provide a well-written modern exposition as a reference of the proof or give one in an answer?

The papers I read also claimed that the proof was simpler in the language of elliptic curves but I am not yet familiar with much theory concerning them so I would very much prefer an elementary proof.

Also, just as a pre-emptive comment I do not consider this question and answer to be a duplicate since the paper they ended up accepting is about as far away from elementary as can be since it was trying to prove a more general statement.

Isky Mathews
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  • I feel like maybe I am missing some important context here, but an arithmetic progression sits on a line, while $x^3$ has a strictly increasing first derivative on positive numbers, so the three cubes can not be all positive. But if we accept negative numbers then we have infinitely many such cubes: $(-n)^3, 0, n^3$. – Cristian Gratie Mar 21 '22 at 22:45
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    @CristianGratie: maybe, but you reasoning also applies to squares, yet there are infinitely many triplets of squares in arithmetic progression (starting with $1,25,49$). – Aphelli Mar 21 '22 at 22:47
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    @CristianGratie: the cubes are all supposed to be positive in the statement of the theorem. I think your comment proves that no three cubes of consecutive integers can be in arithmetic progression but we are potentially considering cubes that are very far apart... – Isky Mathews Mar 21 '22 at 22:52
  • @Mindlack My bad, I guess I'm tired. Cubes in an arithmetic progression would not sit on a line (if seen on a plot of $x^3$) they would simply be equidistant on the $y$ axis. The question makes a lot more sense now. – Cristian Gratie Mar 21 '22 at 22:52
  • It's equivalent to solving $x^3+y^3=2z^3$, so reference in this answer should do. Then it was supposedly solved already by Euler in L. Euler, Algebra 2 (1770), Art. 247; French transl., 2 (1774) 355-360; Opera Omnia I (1) 491 – Sil Mar 21 '22 at 23:04
  • Apologies for the late reply. I believe the answer is fine for my purposes though it's rather unfortunate that there wouldn't be simpler solution (though I guess there is not much reason to suspect there would be). Quite a tough thing to read through! (the book has a chapter 2 with 80+ pages!) – Isky Mathews Mar 22 '22 at 19:48
  • Thank you regardless @Sil. – Isky Mathews Mar 22 '22 at 19:48

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