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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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see the similar [ math.stackexchange.com/questions/93817/sum-of-two-cubes ] – janmarqz Feb 3 '14 at 18:17
    
@janmarqz That question includes irrational numbers – qwr Feb 3 '14 at 18:18
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“I have seen some good proofs, but they are quite long (more than a page) or use many variables.” Welcome to mathematics! – Carsten S Feb 3 '14 at 18:20
    
very simple proof would be to say that i believe it has not solution – dato datuashvili Feb 3 '14 at 18:29
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@Lucian Well, that even seems to fit a book's margin. – Hagen von Eitzen Feb 3 '14 at 18:42
up vote 25 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 $$ r^3 = 18w = 54f (e^2 − f^2) = 54f (e + f) (e − f) = 3^3 \times 2f (e + f) (e − f). $$ 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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Can you present the proof all together? – qwr Feb 12 '14 at 20:24
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I incorporated a proof, using a simple lemma: fermatslasttheorem.blogspot.com/2005/05/… – Yiorgos S. Smyrlis Feb 13 '14 at 9:42
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Incredible! I thoroughly enjoyed this. Thank you – Patrick Shambayati Apr 3 '15 at 1:32
    
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 – Ant Apr 27 at 7:53

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.

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Very unique approach. Can you clarify in under 100 pages? – qwr Feb 12 '14 at 20:15
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@qwr: I have given a reference – user88576 Feb 12 '14 at 21:05
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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 '14 at 21:11
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Although I appreciate the humor of the answer, it really doesn't make any attempt to answer the question. – user61527 Feb 13 '14 at 18:19
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Humour — particularly patronizing humour — does not an appropriate answer make. – Kieren MacMillan Apr 30 '14 at 0:06

Without loss of generality, consider three positive distinct coprime integers $x <y <z$ and $$x^3 + y^3 = z^3$$, First consider the case where $3$ doesn’t divide $xyz$, then the simplest so elegant proof would be by using the general binomial theorem as the following:

$$(x + y - z)^3 = 3(x + y)(z - x)(z – y) + x^3 +y^3 - z^3$$, which can be reduced to: $$(x + y - z)^3 = 3(x + y)(z - x)(z – y) $$, by little Fermat’s theorem we have $(x +y –z)$ is divisible by say $3^k$, where $k$ is positive integer, but we have $(x + y)(z - x)(z – y)$ isn’t divisible by $3$, since they are some factors of $xyz$, because $$x^3 + y^3 = (x + y)(x^2 - xy + y^2) = z^3$$, therefore all prime factors of $(x +y )$ belong to $z$, similarly all prime factors of $(z -x)$ belong to $y$, and all prime factors of $(z -y)$ belong to $x$, but $xyz$ isn't divisible by $3$, so we left out of $3^{3k}$ on the left side of the reduced equation, where as we have a prime factor $3^1$ only on the right side of the reduced equation, therefore this equation is impossible in integers,

Now, can we develop this principle where $3$ divides $xyz$!

To start generalization of this simple obvious result, assume $3^n$ divides $xyz$, where $n \neq 3k – 1$, then the same mentioned proof would be still valid rigorously (that is by equating power of $3$ on both sides of the reduced equation, then they aren’t of equal exponent -

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I'm quite surprised not even a single comment!, I'm afraid my answer isn't clear, but I can explain any query if you like! – bassam karzeddin Apr 19 at 17:40
    
Luckily, this time at MSE, I didn't get so many down votes or deleted, I'm really quite surprised!, also I'm trying hard to reduce this proof much farther, but I'm afraid this wouldn't last for long!, who knows there may be so many generalization too!, you might wonder why I'm talking to my self like that, yes this same answer was down voted at another mathematical community "Quora" quora.com/How-did-Fermat-prove-his-last-theorem-for-n-3/answer/… – bassam karzeddin Apr 20 at 6:41
    
I have made it cleaner to a school student, also I specify it farther to nonzero integers, any little doubt about this case of most elegant proof that isn't even in wiki answers, can be immediately clarified!, also see how this is even much shorter proof than Euler's elementary proof – bassam karzeddin Apr 21 at 6:39

$$x^3+y^3=z^3$$ $$(x+y)(x^2-xy+y^2)=z^3$$ $$z=x+y$$ $$z^2=x^2-xy+y^2$$ $$x^2+2xy+y^2=z^2$$ $$-3xy=0.$$ Only if one of $x$ or $y$ or $z$ equal to $0$, the equation has solution, otherwise it has no solution.

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This answer is incorrect. If $ab=cd$, it does not follow that $a=c$ and $b=d$. (For you, $a=(x+y)$, $b=(x^2-xy+y^2)$, $c=z$, $d=z^2$.) – Zev Chonoles Apr 6 '15 at 7:47

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