[proofplan] We prove the integer statement by reducing any nonzero integer counterexample to a positive-integer counterexample. Once the signs are separated, the equation always says that the sum of two positive cubes is a positive cube. This contradicts [Fermat's Last Theorem](/page/Fermat%27s%20Last%20Theorem) applied with exponent $3$. [/proofplan]

[step:Convert a nonzero integer solution into a positive solution]Assume, for contradiction, that nonzero integers $x,y,z \in \mathbb{Z}$ satisfy \begin{align*} x^3+y^3=z^3. \end{align*} Set \begin{align*} X:=|x|, \qquad Y:=|y|, \qquad Z:=|z|. \end{align*} Since $x,y,z$ are nonzero, $X,Y,Z$ are positive integers. If $x$ and $y$ have the same sign, then $z$ has that same sign, and replacing all three variables by their absolute values gives \begin{align*} X^3+Y^3=Z^3. \end{align*} If $x$ and $y$ have opposite signs, then one term can be moved to the other side. For instance, if $x>0$ and $y<0$, the equation is \begin{align*} X^3-Y^3=z^3. \end{align*} If $z>0$, then \begin{align*} Y^3+Z^3=X^3, \end{align*} while if $z<0$, then \begin{align*} X^3+Z^3=Y^3. \end{align*} The case $x<0$ and $y>0$ is the same with $X$ and $Y$ interchanged. Therefore any nonzero integer solution produces positive integers $A,B,C \in \mathbb{N}$ satisfying \begin{align*} A^3+B^3=C^3. \end{align*}[/step]

[guided]We first remove the sign issue. Define \begin{align*} X:=|x|, \qquad Y:=|y|, \qquad Z:=|z|. \end{align*} These are positive integers because the theorem assumes $x,y,z$ are all nonzero. If $x$ and $y$ have the same sign, then $x^3+y^3$ has that sign, so $z^3$ and $z$ have the same sign as well. Taking absolute values preserves the equation and gives \begin{align*} X^3+Y^3=Z^3. \end{align*} If $x$ and $y$ have opposite signs, then the equation is a difference of two positive cubes equal to the third signed cube. Suppose $x>0$ and $y<0$. Then \begin{align*} X^3-Y^3=z^3. \end{align*} When $z>0$, this rearranges to \begin{align*} Y^3+Z^3=X^3. \end{align*} When $z<0$, it rearranges to \begin{align*} X^3+Z^3=Y^3. \end{align*} The remaining sign pattern $x<0$ and $y>0$ is obtained by interchanging the roles of $X$ and $Y$. Thus in every case a nonzero integer counterexample gives positive integers $A,B,C \in \mathbb{N}$ with \begin{align*} A^3+B^3=C^3. \end{align*}[/guided]

[step:Apply Fermat's Last Theorem with exponent $3$]By [Fermat's Last Theorem](/page/Fermat%27s%20Last%20Theorem), for every integer $n>2$ there are no positive integers $A,B,C \in \mathbb{N}$ satisfying \begin{align*} A^n+B^n=C^n. \end{align*} The exponent $n=3$ satisfies $3>2$, so the positive solution obtained in the previous step is impossible. This contradiction proves that no nonzero integers $x,y,z \in \mathbb{Z}$ satisfy \begin{align*} x^3+y^3=z^3. \end{align*}[/step]

[guided]The previous step reduced the integer statement to the positive-integer equation \begin{align*} A^3+B^3=C^3. \end{align*} We now invoke [Fermat's Last Theorem](/page/Fermat%27s%20Last%20Theorem), which states that if $n>2$, then no positive integers $A,B,C \in \mathbb{N}$ satisfy \begin{align*} A^n+B^n=C^n. \end{align*} Here $n=3$, and $3>2$, so the theorem applies directly. Therefore the positive integers $A,B,C$ constructed from the assumed integer counterexample cannot exist. The assumed nonzero integer solution also cannot exist, completing the proof.[/guided]

<!-- deprecated Eisenstein descent removed by review fix

[step:Introduce the Eisenstein integer arithmetic used in the descent]Let $\omega \in \mathbb{C}$ denote the complex number \begin{align*} \omega = \frac{-1+\sqrt{-3}}{2}, \end{align*} so that $\omega^2+\omega+1=0$ and $\omega^3=1$. Define the Eisenstein integer ring \begin{align*} R := \mathbb{Z}[\omega] = \{a+b\omega : a,b \in \mathbb{Z}\} \subset \mathbb{C}. \end{align*} Define the norm map \begin{align*} N: R &\to \mathbb{Z}_{\ge 0} \\ a+b\omega &\mapsto (a+b\omega)(a+b\omega^2)=a^2-ab+b^2. \end{align*} The norm is multiplicative because complex conjugation is multiplicative: \begin{align*} N(\alpha\beta)=N(\alpha)N(\beta) \qquad \text{for all } \alpha,\beta \in R. \end{align*} The ring $R$ is Euclidean with respect to $N$, hence it is a unique factorization domain. The units of $R$ are precisely \begin{align*} \{\pm 1,\pm \omega,\pm \omega^2\}, \end{align*} and \begin{align*} 3=-\omega^2(1-\omega)^2. \end{align*} We write $\lambda := 1-\omega$.[/step]

[guided]The auxiliary ring is the Eisenstein integer ring. We define \begin{align*} R := \mathbb{Z}[\omega] = \{a+b\omega : a,b \in \mathbb{Z}\}, \end{align*} where \begin{align*} \omega = \frac{-1+\sqrt{-3}}{2}. \end{align*} This choice gives $\omega^2+\omega+1=0$ and $\omega^3=1$, so $\omega$ is a primitive cube root of unity. The relevant arithmetic invariant is the norm map \begin{align*} N: R &\to \mathbb{Z}_{\ge 0} \\ a+b\omega &\mapsto (a+b\omega)(a+b\omega^2)=a^2-ab+b^2. \end{align*} Because complex conjugation sends $a+b\omega$ to $a+b\omega^2$ and respects multiplication, the norm satisfies \begin{align*} N(\alpha\beta)=N(\alpha)N(\beta) \end{align*} for all $\alpha,\beta \in R$. The standard Euclidean division argument for the hexagonal lattice $\mathbb{Z}[\omega]$ shows that $R$ is Euclidean with respect to $N$; therefore $R$ is a unique factorization domain. The units are exactly the elements of norm $1$, namely \begin{align*} \{\pm 1,\pm \omega,\pm \omega^2\}. \end{align*} Finally, direct multiplication gives \begin{align*} (1-\omega)^2=1-2\omega+\omega^2=-3\omega, \end{align*} so, with $\lambda := 1-\omega$, \begin{align*} 3=-\omega^2\lambda^2. \end{align*} This is the special factorization of the rational prime $3$ that drives the descent.[/guided]

[step:Reduce a hypothetical solution to a primitive one]Assume, for contradiction, that nonzero integers $x,y,z \in \mathbb{Z}$ satisfy \begin{align*} x^3+y^3=z^3. \end{align*} Among all such solutions choose one with $|z|$ minimal. Dividing by the common divisor $\gcd(x,y,z)$ if necessary gives a solution with \begin{align*} \gcd(x,y,z)=1. \end{align*} Indeed, if a prime $p$ divides two of $x,y,z$, then the equation forces $p$ to divide the third; hence the common divisor can be removed. Thus the chosen solution is primitive. The congruence of cubes modulo $9$ is \begin{align*} a^3 \equiv 0,1,-1 \pmod 9 \qquad \text{for every } a \in \mathbb{Z}. \end{align*} Since $x^3+y^3=z^3$, exactly one of $x$ and $y$ is divisible by $3$, and $3 \nmid z$. After interchanging $x$ and $y$ if necessary, assume \begin{align*} 3 \mid y, \qquad 3 \nmid xz. \end{align*}[/step]

[guided]Suppose a counterexample exists. We choose one for which $|z|$ is as small as possible among all counterexamples. This is possible because $|z|$ is a positive integer. First we make the solution primitive. If a prime $p$ divides two of $x,y,z$, then the equation \begin{align*} x^3+y^3=z^3 \end{align*} forces $p$ to divide the remaining one. For example, if $p \mid x$ and $p \mid y$, then $p \mid z^3$, hence $p \mid z$. The other two cases are identical after moving terms. Therefore any common factor can be divided out, producing a smaller solution unless the solution is already primitive. Thus we may assume \begin{align*} \gcd(x,y,z)=1. \end{align*} Next we record the role of the prime $3$. For every integer $a$, reducing the residues $0,1,\dots,8$ modulo $9$ gives \begin{align*} a^3 \equiv 0,1,-1 \pmod 9. \end{align*} If neither $x$ nor $y$ were divisible by $3$, then $x^3$ and $y^3$ would each be congruent to $1$ or $-1$ modulo $9$. Their sum would be congruent to $2,0,$ or $-2$ modulo $9$. Since $z^3$ is congruent only to $0,1,$ or $-1$ modulo $9$, the only possible case would be $x^3+y^3 \equiv 0 \pmod 9$, forcing $3 \mid z$. Then primitivity would fail. If both $x$ and $y$ were divisible by $3$, primitivity would also fail. Hence exactly one of $x$ and $y$ is divisible by $3$. Interchanging the two variables does not change the equation, so we assume \begin{align*} 3 \mid y, \qquad 3 \nmid xz. \end{align*}[/guided]

[step:Factor the equation in $\mathbb{Z}[\omega]$ and isolate a cube] In $R$ we have \begin{align*} z^3=x^3+y^3=(x+y)(x+\omega y)(x+\omega^2y). \end{align*} The three factors on the right are pairwise coprime in $R$ except for possible factors associated to $\lambda=1-\omega$. Since $3 \mid y$ and $3 \nmid x$, the element $x+\omega y$ is not divisible by $\lambda$: modulo $\lambda$, we have $\omega \equiv 1$, so \begin{align*} x+\omega y \equiv x+y \equiv x \not\equiv 0 \pmod{\lambda}. \end{align*} Thus $x+\omega y$ is coprime in $R$ to both $x+y$ and $x+\omega^2y$. Because $R$ is a unique factorization domain and the product of these pairwise coprime factors is the cube $z^3$, every prime exponent in the factorization of $x+\omega y$ is divisible by $3$. Hence there are a unit $u \in \{\pm1,\pm\omega,\pm\omega^2\}$ and elements $a,b \in \mathbb{Z}$ such that \begin{align*} x+\omega y = u(a+b\omega)^3. \end{align*} Multiplying $a+b\omega$ by a unit if necessary absorbs $u$, because every unit of $R$ is itself a cube up to sign and sign is also a cube. Therefore we may write \begin{align*} x+\omega y=(a+b\omega)^3 \end{align*} for some $a,b \in \mathbb{Z}$.[/step]

[guided]The point of passing to $\mathbb{Z}[\omega]$ is that the sum of two cubes factors completely. In $R$ we have \begin{align*} x^3+y^3=(x+y)(x+\omega y)(x+\omega^2y), \end{align*} so the equation becomes \begin{align*} z^3=(x+y)(x+\omega y)(x+\omega^2y). \end{align*} We want to conclude that the middle factor $x+\omega y$ is itself a cube up to multiplication by a unit. Unique factorization permits that conclusion once we know this factor is coprime to the other two factors. The only possible common divisor among these factors lies above the rational prime $3$, represented in $R$ by $\lambda=1-\omega$. This is because subtracting any two of the factors gives a multiple of $(1-\omega)y$ or $(1-\omega^2)y$, and primitivity rules out rational primes dividing both $x$ and $y$. We now check that even $\lambda$ does not divide $x+\omega y$. Modulo $\lambda$, the relation $1-\omega=0$ gives $\omega \equiv 1$. Hence \begin{align*} x+\omega y \equiv x+y \pmod{\lambda}. \end{align*} Since $3 \mid y$ and $3 \nmid x$, we have $x+y \equiv x \not\equiv 0 \pmod 3$, and therefore $x+\omega y$ is not divisible by $\lambda$. Now $R$ is a unique factorization domain. The product \begin{align*} (x+y)(x+\omega y)(x+\omega^2y)=z^3 \end{align*} is a cube, and the factor $x+\omega y$ is coprime to the other two factors. Therefore every irreducible divisor of $x+\omega y$ occurs with exponent divisible by $3$. This gives a unit $u \in \{\pm1,\pm\omega,\pm\omega^2\}$ and integers $a,b \in \mathbb{Z}$ such that \begin{align*} x+\omega y=u(a+b\omega)^3. \end{align*} Multiplying $a+b\omega$ by a suitable unit only changes its cube by a unit, and $-1=(-1)^3$ is itself a cube. Thus the unit can be absorbed, and we may write \begin{align*} x+\omega y=(a+b\omega)^3. \end{align*}[/guided]

[step:Expand the Eisenstein cube and extract divisibility]Expanding and using $\omega^2=-1-\omega$ gives \begin{align*} (a+b\omega)^3 &=a^3+3a^2b\omega+3ab^2\omega^2+b^3 \\ &=a^3+b^3-3ab^2+\omega(3a^2b-3ab^2). \end{align*} Comparing coefficients in the basis $\{1,\omega\}$ gives \begin{align*} y=3ab(a-b). \end{align*} Since $3 \mid y$, this is compatible with the previous step. Moreover the three integers $a$, $b$, and $a-b$ are pairwise coprime up to factors of $3$; primitivity of $(x,y,z)$ and $3 \nmid x$ force exactly one of them to be divisible by $3$. Consequently \begin{align*} \frac{y}{3}=ab(a-b) \end{align*} is a product of three pairwise coprime integers whose product is a cube up to sign. Hence each factor is a cube up to sign. Thus there exist integers $r,s,t \in \mathbb{Z}$ such that \begin{align*} a=r^3, \qquad b=s^3, \qquad a-b=t^3. \end{align*} Therefore \begin{align*} r^3-s^3=t^3. \end{align*} Equivalently, \begin{align*} s^3+t^3=r^3. \end{align*}[/step]

[guided]We now translate the statement that $x+\omega y$ is an Eisenstein cube back into ordinary integer equations. Expanding gives \begin{align*} (a+b\omega)^3 &=a^3+3a^2b\omega+3ab^2\omega^2+b^3. \end{align*} Since $\omega^2=-1-\omega$, this becomes \begin{align*} (a+b\omega)^3 &=a^3+b^3-3ab^2+\omega(3a^2b-3ab^2). \end{align*} Comparing the coefficient of $\omega$ with $x+\omega y$ gives \begin{align*} y=3a^2b-3ab^2=3ab(a-b). \end{align*} So \begin{align*} \frac{y}{3}=ab(a-b). \end{align*} The coprimality information comes from the primitive nature of the original solution. Any prime divisor common to two of $a$, $b$, and $a-b$ divides the third relevant difference as well, and through the identities above it would force a common divisor of $x$ and $y$, except possibly for the prime $3$. The congruence condition $3 \nmid x$ rules out the uncontrolled appearance of $3$ in more than one of the three factors. Thus $a$, $b$, and $a-b$ are pairwise coprime after accounting for the single forced factor of $3$ already separated in $y=3ab(a-b)$. Because $x+\omega y$ is a cube in the unique factorization domain $R$, its norm \begin{align*} N(x+\omega y)=x^2-xy+y^2 \end{align*} is an ordinary integer cube. The factorization above then implies that the pairwise coprime integer factors $a$, $b$, and $a-b$ each have all prime exponents divisible by $3$. Therefore there exist integers $r,s,t \in \mathbb{Z}$ such that \begin{align*} a=r^3, \qquad b=s^3, \qquad a-b=t^3. \end{align*} Substituting $a-b=t^3$ gives \begin{align*} r^3-s^3=t^3, \end{align*} which is the same as \begin{align*} s^3+t^3=r^3. \end{align*} This is a new solution of the same Fermat equation.[/guided]

[step:Contradict minimality by producing a smaller solution]The new solution \begin{align*} s^3+t^3=r^3 \end{align*} has nonzero integer entries. Also \begin{align*} |r|^3=|a| < N(a+b\omega)=|z|, \end{align*} after replacing by associates if necessary in the preceding cube representation. Hence \begin{align*} 0<|r|<|z|. \end{align*} This contradicts the minimal choice of $|z|$. Therefore no nonzero integers $x,y,z \in \mathbb{Z}$ satisfy \begin{align*} x^3+y^3=z^3. \end{align*}[/step]

[guided]The previous step produced another integer solution \begin{align*} s^3+t^3=r^3. \end{align*} It remains to check that this really contradicts the way we chose the original solution. The entries are nonzero, because if one of $r,s,t$ were zero, then one of $a$, $b$, or $a-b$ would be zero, forcing $y=3ab(a-b)=0$, contrary to the theorem's nonzero hypothesis. The descent is strict because $a+b\omega$ is a proper cube root of $x+\omega y$ in the norm. Since \begin{align*} x+\omega y=(a+b\omega)^3, \end{align*} we have \begin{align*} N(x+\omega y)=N(a+b\omega)^3. \end{align*} The norm is positive on nonzero Eisenstein integers, so \begin{align*} N(a+b\omega)<N(x+\omega y) \end{align*} whenever $N(a+b\omega)>1$. The construction of $r$ from the integer factor $a=r^3$ gives, after choosing associates in the cube representation, the strict bound \begin{align*} 0<|r|<|z|. \end{align*} Thus $s^3+t^3=r^3$ is a nonzero integer solution whose right-hand side has smaller absolute value than the chosen minimal counterexample. This contradiction proves that no counterexample exists.[/guided]

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