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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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    $\begingroup$ “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
  • $\begingroup$ When the exponent is a prime, use & abuse Newton's binomial theorem. All the terms of any row of Pascal's triangle whose index is a prime number divide through it. This is what ultimately makes the Diophantine equations $a^p+b^p=c^p$ impossible, starting with $p=3$. It's a fairly simple (but very laborious) question of exploiting divisibility. With all primes $>2$ out of the way, their multiples soon follow, so all that's left is $n=4$. You inevitably arrive at a point where the product of two terms equals a product of n terms. For $n=2$ this works, but not for $n=p>2$, due to co-primality. $\endgroup$ – Lucian Feb 3 '14 at 18:38
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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
  • $\begingroup$ @Lucian Can you add a bit more detail and post that as an answer? $\endgroup$ – qwr Feb 3 '14 at 18:50
  • $\begingroup$ Unfortunately, as I've already said, my proof for $n=3$ is also quite long (though by no means complex), and it does use quite a few extra variables, so... $\endgroup$ – Lucian Feb 4 '14 at 3:23
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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.

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  • $\begingroup$ 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
  • $\begingroup$ 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
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There’s a wonderful elementary (and fairly short) proof in this paper by S.Dolan.

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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

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  • $\begingroup$ 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

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