# Prove that $x^3 + y^3 = z^3$ has no integer solutions as briefly as possible

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
“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
@Lucian Well, that even seems to fit a book's margin. – 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 $$r^3 = 18w = 54f (e^2 − f^2) = 54f (e + f) (e − f) = 33 \times 2f (e + f) (e − f).$$ 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

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Can you present the proof all together? – qwr Feb 12 '14 at 20:24
I incorporated a proof, using a simple lemma: fermatslasttheorem.blogspot.com/2005/05/… – Yiorgos S. Smyrlis Feb 13 '14 at 9:42
Incredible! I thoroughly enjoyed this. Thank you – user602819 Apr 3 '15 at 1:32

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
@qwr: I have given a reference – user88576 Feb 12 '14 at 21:05
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
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
Humour — particularly patronizing humour — does not an appropriate answer make. – Kieren MacMillan Apr 30 '14 at 0:06

$$x^3+y^3=z^3$$ $$(x+y)(x^2-xy+y^2)=z^3$$ $$z=x+y$$ $$z^2=x^2-xy+y^2$$ $$x^2+2xy+y^2=z^2$$ $$-3xy=0.$$ 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