Can someone provide the proof of the special case of Fermat's Last Theorem for $n=3$, i.e., that $$ x^3 + y^3 = z^3, $$ has no positive integer solutions, as briefly as possible?
I have seen some good proofs, but they are quite long (longer than a page) or use many variables. However, I would rather have an elementary long proof with many variables than a complex short proof.
Edit. Even if the bounty expires I will award one to someone if they have a satisfying answer.
On
Eight years and one elementary number theory course later, I can give a brief but not as elementary sketch of an answer based on the ring of Eisenstein integers. They have a strong connection to cubes, considering the cubic root of unity $\omega$. Assume it is already established the Eisenstein integers $\mathbb Q(\omega)$ are a Euclidean domain, so UFD. This answer follows that in Dörrie's 100 great problems of elementary mathematics, but he spends some pages reproving results about Euclidean domains.
First off we can assume in $\alpha^3 + \beta^3 = \gamma^3$ that $\alpha, \beta, \gamma$ are coprime in the Eisenstein integers due to unique factorization. By trivially replacing $\gamma$ with $-\gamma$ the equation is $\alpha^3 + \beta^3 + \gamma^3 = 0$.
Now $i\sqrt 3 = \omega - \omega^2 = 2 \omega + 1$ is an Eisenstein prime by properties of norm. (Dörrie calls this $J-O$, where he defines $J = - \omega^2 = (1 + i \sqrt 3)/2, O = - \omega = (1-i \sqrt 3)/2$. He uses $\{J, O\}$ as a basis instead of $\{1,\omega\}$ or $\{\omega, \omega^2\}$ for some reason?)
By some basic computations and casework of $a,b$ mod 3 I'm too lazy to write up, if 3 divides $a + b$ then $i \sqrt 3$ divides $a + b \omega$, and more calculations show Lemma: if $i \sqrt 3$ doesn't divide $a + b\omega$, then $(a + b\omega)^3 \equiv \pm 1 \pmod 9$.
Now Lemma: $\alpha^3 + \beta^3 + \gamma^3 = 0$ implies $i \sqrt 3$ divides exactly one of $\alpha, \beta, \gamma$. This holds because suppose $i \sqrt 3$ didn't divide any of $\alpha, \beta, \gamma$. Then $\alpha^3 \equiv e, \beta^3 \equiv f, \gamma^3 \equiv g \pmod 9$ where $e^2 = f^2 = g^2 = 1$. But by casework mod 9, this is impossible. And we know $\alpha, \beta, \gamma$ are coprime so can't share a factor of $i \sqrt 3$.
Finally, $\alpha \beta \gamma \ne 0$ is impossible by some arguments of common divisor and infinite descent on the norm given in R. Andrew Ohana's more algebraic proof. Actually I should've just linked his proof instead of writing all the above, since the algebraic structure is more enlightening. But I'll leave the answer as a source of references and getting down ideas.
Main idea. The proof that follows is based on the infinite descent, i.e., we shall show that if $(x,y,z)$ is a solution, then there exists another triplet $(k,l,m)$ of smaller integers, which is also a solution, and this leads apparently to a contradiction.
Assume instead that $x, y, z\in\mathbb Z\smallsetminus\{0\}$ satisfy the equation (replacing $z$ by $-z$) $$x^3 + y^3 + z^3 = 0,$$ with $x, y$ and $z$ pairwise coprime. (Clearly at least one is negative.) One of them should be even, whereas the other two are odd. Assume $z$ to be even.
Then $x$ and $y$ are odd. If $x = y$, then $2x^3 = −z^3$, and thus $x$ is also even, a contradiction. Hence $x\ne y$.
As $x$ and $y$ are odd, then $x+y$, $x-y$ are both even numbers. Let $$ 2u = x + y, \quad 2v = x − y, $$ where the non-zero integers $u$ and $v$ are also coprime and of different parity (one is even, the other odd), and $$ x = u + v\quad \text{and}\quad y = u − v. $$ It follows that $$ −z^3 = (u + v)^3 + (u − v)^3 = 2u(u^2 + 3v^2). \tag{1} $$ Since $u$ and $v$ have different parity, $u^2 + 3v^2$ is an odd number. And since $z$ is even, then $u$ is even and $v$ is odd. Since $u$ and $v$ are coprime, then $$ {\mathrm{gcd}}\,(2u,u^2 + 3v^2)={\mathrm{gcd}}\,(2u,3v^2)\in\{1,3\}. $$
Case I. $\,{\mathrm{gcd}}\,(2u,u^2 + 3v^2)=1$.
In this case, the two factors of $−z^3$ in $(1)$ are coprime. This implies that $3\not\mid u$ and that both the two factors are perfect cubes of two smaller numbers, $r$ and $s$. $$ 2u = r^3\quad\text{and}\quad u^2 + 3v^2 = s^3. $$ As $u^2 + 3v^2$ is odd, so is $s$. We now need the following result:
Lemma. If $\mathrm{gcd}\,(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.
Proof. See here.
Thus, if $s$ is odd and if it satisfies an equation $s^3 = u^2 + 3v^2$, then it can be written in terms of two coprime integers $e$ and $f$ as $$ s = e^2 + 3f^2, $$ so that $$ u = e ( e^2 − 9f^2) \quad\text{and}\quad v = 3f ( e^2 − f^2). $$ Since $u$ is even and $v$ odd, then $e$ is even and $f$ is odd. Since $$ r^3 = 2u = 2e (e − 3f)(e + 3f), $$ the factors $2e$, $(e–3f )$, and $(e+3f )$ are coprime since $3$ cannot divide $e$. If $3\mid e$, then $3\mid u$, violating the fact that $u$ and $v$ are coprime. Since the three factors on the right-hand side are coprime, they must individually equal cubes of smaller integers $$ −2e = k^3,\,\,\, e − 3f = l^3,\,\,\, e + 3f = m^3, $$ which yields a smaller solution $k^3 + l^3 + m^3= 0$. Therefore, by the argument of infinite descent, the original solution $(x, y, z)$ was impossible.
Case II. $\,{\mathrm{gcd}}\,(2u,u^2 + 3v^2)=3$.
In this case, the greatest common divisor of $2u$ and $u^2 + 3v^2$ is $3$. That implies that $3\mid u$, and one may express $u = 3w$ in terms of a smaller integer, $w$. Since $4\mid u$, so is $w$; hence, $w$ is also even. Since $u$ and $v$ are coprime, so are $v$ and $w$. Therefore, neither $3$ nor $4$ divide $v$.
Substituting $u$ by $w$ in $(1)$ we obtain $$ −z^3 = 6w(9w^2 + 3v^2) = 18w(3w^2 + v^2) $$ Because $v$ and $w$ are coprime, and because $3\not\mid v$, then $18w$ and $3w^2 + v^2$ are also coprime. Therefore, since their product is a cube, they are each the cube of smaller integers, $r$ and $s$: $$ 18w = r^3 \quad\text{and}\quad 3w^2 + v^2 = s^3. $$ By the same lemma, as $s$ is odd and equal to a number of the form $3w^2 + v^2$, it too can be expressed in terms of smaller coprime numbers, $e$ and $f$: $$ s = e^2 + 3f^2. $$ A straight-forward calculation shows that $$ v = e (e^2 − 9f^2) \quad\text{and}\quad w = 3f (e^2 − f^2). $$ Thus, $e$ is odd and $f$ is even, because $v$ is odd. The expression for $18w$ then becomes $$ r^3 = 18w = 54f (e^2 − f^2) = 54f (e + f) (e − f) = 3^3 \times 2f (e + f) (e − f). $$ Since $3^3$ divides $r^3$ we have that $3$ divides $r$, so $(r /3)^3$ is an integer that equals $2f (e + f) (e − f)$. Since $e$ and $f$ are coprime, so are the three factors $2e$, $e+f$, and $e−f$; therefore, they are each the cube of smaller integers, $k$, $l$, and $m$. $$ −2e = k^3,\,\,\, e + f = l^3,\,\,\, e − f = m^3, $$ which yields a smaller solution $k^3 + l^3 + m^3= 0$. Therefore, by the argument of infinite descent, the original solution $(x, y, z)$ was impossible.
Note. See also here.