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.

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    $\begingroup$ see the similar [ math.stackexchange.com/questions/93817/sum-of-two-cubes ] $\endgroup$ – janmarqz Feb 3 '14 at 18:17
  • $\begingroup$ @janmarqz That question includes irrational numbers $\endgroup$ – qwr Feb 3 '14 at 18:18
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    $\begingroup$ “I have seen some good proofs, but they are quite long (more than a page) or use many variables.” Welcome to mathematics! $\endgroup$ – Carsten S Feb 3 '14 at 18:20
  • $\begingroup$ very simple proof would be to say that i believe it has not solution $\endgroup$ – dato datuashvili Feb 3 '14 at 18:29
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    $\begingroup$ @Lucian Well, that even seems to fit a book's margin. $\endgroup$ – Hagen von Eitzen Feb 3 '14 at 18:42

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.

  • $\begingroup$ Can you present the proof all together? $\endgroup$ – qwr Feb 12 '14 at 20:24
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    $\begingroup$ I incorporated a proof, using a simple lemma: fermatslasttheorem.blogspot.com/2005/05/… $\endgroup$ – Yiorgos S. Smyrlis Feb 13 '14 at 9:42
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    $\begingroup$ Incredible! I thoroughly enjoyed this. Thank you $\endgroup$ – Patrick Shambayati Apr 3 '15 at 1:32
  • $\begingroup$ But the infinite descent only works for positive integers.. You should show that there is a positive solution which gets progressively smaller..so at least one of the positive solution of $(x,y,z)$ gets smaller.. It is not immediately clear to me why is it so $\endgroup$ – Ant Apr 27 '16 at 7:53

Assume $x^3+y^3=z^3$, $x$, $y$, $z$ c0-prime, $x$ even, $y$ odd, $z$ odd.

Let $x+y=z+d$. Cubing the latter, subtracting the hypothesis: $$3x^2y+3y^2x=3z^2d+3d^2z+d^3.$$ Transposing and substituting: $$3(z+d)(xy-zd)=d^3.$$ $3\mid d$, so let $d=3e$. The equation becomes: $$3(z+3e)(xy-3ze)=27e^3$$. Dividing by $3$, $$(z+3e)(xy-3ze)=9e^3.$$ The left side is divisible by $e^3$. However, each factor cannot be divisible by $e$, or else $e\mid(z+3e)$, or $e\mid $z, and $e\mid (xy 3ze)$, or $e \mid xy$. Hence $e\mid(z\text{ and }x)$ or $e \mid (z\text{ and }y)$, which is impossible because co-prime.

$e\ne1$ because we would have $(z+3)(xy-3z)=9$.

$9$ divides the left side, but both factors cannot be divisible by $3$,or $3 \mid z$ and $3 \mid xy$, impossible.

Returning to $$(z+3e)(xy-3ze)=9e^3,$$ we repeat the argument just completed to get the result:
the 2 factors on the left can only be divisible be $e^3$, $9e^3$, $1$, and $9$,so we have the $4$ possibilities:
(a) $z+3e=9e^3$, $(xy-3ez)=1$;
(b) $z+3=e^3$, $(xy-3ez)=9$
(c) $(z+3e)=9$, $(xy-3ez)=e^3$,
(d) $(z+3e)=1$, $(xy-3ez)=9e^3$.
Now (d) and (c) are clearly impossible. In (a) and (b), the expression $xy-3ez=xy -dz$ is the difference of 2 even numbers; their difference could not be $1$ or $9$. Q.E.D. Ed Gray

  • $\begingroup$ this is pretty unreadable $\endgroup$ – baxx Dec 18 '16 at 21:40
  • $\begingroup$ As haxx said in the above comment, the original revision was very difficult to read. I have at least added some paragraphs and MathJax (which can be considered the bare minimum). If you can see further improvements, please, edit your post further. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Mar 14 '17 at 7:05
  • $\begingroup$ This answer is simply wrong. You claim that $(z+3)(xy-3z) = 9e^3$ is impossible if $x,y,z$ are coprime with $x$ even and $y,z$ odd. But your 'argument' is totally invalid. (It is already invalid for $e = 1$, but by luck you made a true claim.) You have many more than 4 possibilities. A counter-example is not needed to invalidate your argument, but in this case there is one: $(x,y,z,e) = (2,1461,95,10)$. $\endgroup$ – user21820 Mar 14 '17 at 8:32
  • $\begingroup$ The comment by user 21820 used the erroneous equation (z + 3)(xy - 3z). $\endgroup$ – Edwin Gray Mar 15 '17 at 12:09

protected by user99914 Jan 28 '18 at 5:01

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