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This is one of those recreational questions that aren't really about research. I found a curious remark in an old volume of American Mathematical Monthly (1922) which I'll quote below:

In the Proceedings of the Royal Society of Edinburgh, vol.7, p.144, in some mathematical notes by professor P.G. Tait, it is stated:

"If $x^3+y^3=z^3$, then $(x^3+z^3)^3y^3+(x^3-y^3)^3z^3=(z^3+y^3)^3x^3$.

This furnishes an easy proof of the impossibility of finding two integers the sum of whose cubes is a cube."

How does this "easy proof" follow? Students are notoriously suspicious of those steps which an author announces as "easy", and are sometimes inclined to believe that the word is used in a humorous sense. [...] There are of course proofs in existence that the sum of two cubes cannot be a cube.

Did anyone manage to find a proof of FLT for the exponent 3 using this identity or is the alluded proof another illusion that did not fit in the margin?

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Ribenboim's book "Fermat's Last Theorem for Amateurs" mentions this proof in a list of published proofs for exponent $3$, but gives no details. Unfortunately, the Proceedings of the Royal Society of Edinburgh doesn't seem to be online before volume 78 (at least I can't find it), so I don't know if the context for this statement gives any clues. –  Henry Cohn Feb 2 '12 at 0:12
Mordell-Faltings is for curves of genus $2$ or more. The Fermat cubic has genus $1$. The map $(x,y,x) \mapsto ((z^3+z^3) y, (x^3-y^3)z, (y^3+z^3)x)$ has degree $4$, so must be multiplication by $2$ (up to translation by a rational torsion point). This suggests a proof by $2$-descent, but such a proof, though possible for this curve (the $2$-part of Sha is trivial), is not nearly so easy that exhibiting the map is tantamount to a solution. –  Noam D. Elkies Feb 2 '12 at 1:10
Indeed, in characteristic zero a curve that admits a self-map of degree $>1$ must have genus $0$ or $1$ (a consequence of the Riemann-Hurwitz formula), so we can never get to use Mordell-Faltings this way to effectively list all the rational points. –  Noam D. Elkies Feb 2 '12 at 3:15
For those interested in seeing the original: everything is digitised these days: archive.org/stream/proceedingsroya40edingoog#page/n166/mode/2up - and contrary to hope of Henry, there are no clues. –  Vladimir Dotsenko Feb 2 '12 at 19:01
On an historical side, the formula mentioned by Tait was known before. See for example the Cauchy-Desboves formulas en.wikipedia.org/wiki/… I believe this formula was even mentioned (in an equivalent form) by Bachet in his translation of the Arithmetica of Diophantus. See archive.org/stream/oeuvresdefermat963ferm#page/246/mode/2up –  François Brunault Feb 2 '12 at 19:34
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1 Answer

If $x^3 + y^3 = z^3$ then it is easy to prove that without loss of generality a co-prime version must exist such that $x, y, z$ have no common factor other than 1. Further it is easy to prove that there must be two co-prime positive numbers $p$ and $q$ such that one of the three, say $z$, has the form $z^3 = 2p(p^2 + 3q ^2)$ with $p$ and $q$ of opposite parity (one odd, the other even), therefore $p^2 + 3q^2$ being an odd cube and $2p$ being an even cube, and gcd $(2p, p^2 + 3q^2)$ being 1 or 3. As Euler has shown, it is possible then to find a smaller solution to Fermat's problem for $n$ = 3. The proof of the last step is somewhat tedious. In both cases it includes the lemma that there exist two co-prime numbers $a, b$ of opposite parity such that $p = a^3 - 9ab^2$.

Perhaps Tait thought to use the result of his equation (a), "every cube is the difference of two squares, one at least of which is divisible by 9" in order to easen this last step.

Concerning part (b), it is obvious that Tait's second equation has the same structure as the original one, but now containing three cubes that are definitely greater than the original cubes.

$((x^3 + z^3)y)^3 + ((x^3 - y^3)z)^3 = ((z^3+y^3)x)^3$

This procedure of increasing the cubes can be repeated in infinity. Although it is not justified to reverse the derivation and to conclude that by this method every triple of cubes could be reduced to smaller cubes, it cannot be excluded that Tait was misled by this slip.

I don't believe that there is an "easy" proof, and I have to offer three reasons:

The first one (and admittedly the weakest) is that I have not suceeded in finding an easy solution. Any simplification like

$(2x^3 + y^3)^3y^3 + (x^3 - y^3)^3(x^3+y^3) = (x^3 + 2y^3)^3x^3$

$((2x^3 + y^3)^3 + (x^3 - y^3)^3)y^3 = ((2y^3 + x^3)^3 + (y^3 - x^3)^3)x^3$

leads easily to the result that the derived equation is correct, finally leaving 0 = 0, but no information about the impossibility of an integer character of the roots appeares obvious to me.

A certainly stronger reason is that this, if correct very interesting, note has, as far as I can see, never been mentioned again by Tait and has not been included in his collected papers, which were edited and furnished with a preface by Tait himself. (The note has however been mentioned in a great many of other sources including Ribenboim's popular book and Wikipedia.)

The third and most important reason however, outweighing all others, is that this question has been asked in the American Mathematical Monthly in 1914, 1916, 1919, 1920, 1921, and 1922, which at that time was edited by eminent mathematicians like Hurwitz and certainly read by many others. Last but not least, the problem has been open here in MathOverflow for more than one year. No easy way could be shown.

So it is very probable that this note belongs to the same category as Fermat's original one.

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Google shows 1170 results for >>"Peter Guthrie Tait (1872)"<< and 213 results for >>"Tait (1872)" cubes<<. –  Rhett Butler Feb 15 '13 at 18:45
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