Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
see the similar [ math.stackexchange.com/questions/93817/sum-of-two-cubes ] –  janmarqz Feb 3 at 18:17
    
@janmarqz That question includes irrational numbers –  qwr Feb 3 at 18:18
27  
“I have seen some good proofs, but they are quite long (more than a page) or use many variables.” Welcome to mathematics! –  Carsten Schultz Feb 3 at 18:20
    
@CarstenSchultz :P There should be a simple proof somewhere –  qwr Feb 3 at 18:22
7  
@Lucian Well, that even seems to fit a book's margin. –  Hagen von Eitzen Feb 3 at 18:42

2 Answers 2

up vote 16 down vote accepted
+50

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 $$ r^3 = 18w = 54f (e^2 − f^2) = 54f (e + f) (e − f) = 33 \times 2f (e + f) (e − f). $$ Since $33$ 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. A simple proof of Fermat's Last Theorem when $n=3$ can be found here:

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html

share|improve this answer
    
Can you present the proof all together? –  qwr Feb 12 at 20:24
1  
I incorporated a proof, using a simple lemma: fermatslasttheorem.blogspot.com/2005/05/… –  Yiorgos S. Smyrlis Feb 13 at 9:42

By a theorem of A. Wiles (Ann. Math. 142) we have that $x^n+y^n=z^n$ has no solutions for $n\geq 3$. Now put $n=3$.

Q. E. D.

share|improve this answer
8  
Very unique approach. Can you clarify in under 100 pages? –  qwr Feb 12 at 20:15
2  
@qwr: I have given a reference –  user88576 Feb 12 at 21:05
11  
I've always appreciated that (unlike on many internet forums) on Math.SE answers actually try to answer the question (as opposed to, say, amusing the audience). This one is obviously an exception... –  Grigory M Feb 12 at 21:11
6  
Although I appreciate the humor of the answer, it really doesn't make any attempt to answer the question. –  user61527 Feb 13 at 18:19
    
Humour — particularly patronizing humour — does not an appropriate answer make. –  Kieren MacMillan Apr 30 at 0:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.