# The equation \$x^3 + y^3 = z^3\$ has no integer solutions

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 lớn someone if they have a satisfying answer.  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, và this leads apparently lớn a contradiction.

Assume instead that \$x, y, zinpacmanx.combb Zsmallsetminus\$ 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 lớn be even.

Then \$x\$ và \$y\$ are odd. If \$x = y\$, then \$2x^3 = −z^3\$, & thus \$x\$ is also even, a contradiction. Hence \$x e y\$.

As \$x\$ và \$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\$ và \$v\$ are also coprime & of different parity (one is even, the other odd), và \$\$x = u + vquad extandquad y = u − v.\$\$It follows that\$\$−z^3 = (u + v)^3 + (u − v)^3 = 2u(u^2 + 3v^2). ag1\$\$Since \$u\$ and \$v\$ have different parity, \$u^2 + 3v^2\$ is an odd number. And since \$z\$ is even, then \$u\$ is even & \$v\$ is odd. Since \$u\$ và \$v\$ are coprime, then\$\$ pacmanx.comrmgcd,(2u,u^2 + 3v^2)=pacmanx.comrmgcd,(2u,3v^2)in1,3.\$\$

Case I. \$,pacmanx.comrmgcd,(2u,u^2 + 3v^2)=1\$.

In this case, the two factors of \$−z^3\$ in \$(1)\$ are coprime. This implies that \$3 otmid u\$ & that both the two factors are perfect cubes of two smaller numbers, \$r\$ và \$s\$.\$\$2u = r^3quad extandquad u^2 + 3v^2 = s^3.\$\$As \$u^2 + 3v^2\$ is odd, so is \$s\$. We now need the following result:

Lemma. If \$pacmanx.comrmgcd,(a,b)=1\$, then every odd factor of \$a^2 + 3b^2\$ has this same form.

Proof. See here.

Thus, if \$s\$ is odd & 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 extandquadv = 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 )\$, và \$(e+3f )\$ are coprime since \$3\$ cannot divide \$e\$. If \$3mid e\$, then \$3mid u\$, violating the fact that \$u\$ & \$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. \$,pacmanx.comrmgcd,(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 \$3mid u\$, và one may express \$u = 3w\$ in terms of a smaller integer, \$w\$. Since \$4mid u\$, so is \$w\$; hence, \$w\$ is also even. Since \$u\$ và \$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\$ và \$w\$ are coprime, và because \$3 otmid 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\$ & \$s\$:\$\$18w = r^3 quad extandquad3w^2 + v^2 = s^3.\$\$By the same lemma, as \$s\$ is odd and equal to lớn a number of the khung \$3w^2 + v^2\$, it too can be expressed in terms of smaller coprime numbers, \$e\$ và \$f\$:\$\$s = e^2 + 3f^2.\$\$A straight-forward calculation shows that\$\$v = e (e^2 − 9f^2) quad extandquad 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 imes 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\$, và \$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.