By Lemma 1.31, we can conclude that \(\left[n \right]^3 = 1 \lor \left[n \right]^3 = 8\).

By hypothesis we know that \(3 \nmid abc\), which implies that \(3 \nmid a\), \(3 \nmid b\) and \(3 \nmid c\).
By repeatedly applying Lemma 3.1 for each case, we can conclude that

By hypothesis we have that \(3 \mid a^3 + b^3 = c^3\), which implies that \(3 \mid c\), from which we can conclude that \(3 \mid \gcd (a,b,c)\).

By hypothesis we have that \(3 \mid c^3 - a^3 = b^3\), which implies that \(3 \mid b\), from which we can conclude that \(3 \mid \gcd (a,b,c)\).

By hypothesis we have that \(3 \mid c^3 - b^3 = a^3\), which implies that \(3 \mid a\), from which we can conclude that \(3 \mid \gcd (a,b,c)\).

By contradiction we assume that

By Lemma 1.30 we can assume that \(\gcd (a,b,c)=1\).
By Theorem 3.2 we can assume that \(3 \mid a b c\), from which it follows that

We proceed by analysing each case:

• Case \(3 \mid a\).
  Let \(a'=-c\), \(b'=b\), \(c'=-a\), then \(3 \mid c'\) and

  \[ (a'\neq 0) \land (b'\neq 0) \land (c' \neq 0). \]

  Then \(3 \nmid a'\) since otherwise by Lemma 3.4 we would have that \(3 \mid \gcd (a,b,c)=1\) which is absurd.
  Analogously, by Lemma 3.3 we have that \(3 \nmid b'\).
  By contradiction we assume that \(\gcd (a',b') \neq 1\) which, by basic divisibility properties, implies that there is a prime \(p\) such that \(p \mid a'\) and \(p \mid b'\). It follows that \(p \mid b'^3 + a'^3 = b^3 - c^3 = -a^3\), which implies that \(p \mid a\).
  Therefore \(p \mid \gcd (a,b,c)=1\) which is absurd.
  Moreover, we have that \(a'^3 + b'^3 = -c^3 + b^3 = -a^3 = c'^3\) that contradicts our hypothesis.

• Case \(3 \mid b\).
  Let \(a'=a\), \(b'=-c\), \(c'=-b\).
  The rest of the proof is analogous to the first case using Lemma 3.3 and Lemma 3.5.

• Case \(3 \mid c\). Let \(a'=a\), \(b'=b\), \(c'=c\).
  The rest of the proof is analogous to the first case using Lemma 3.4 and Lemma 3.5.

Therefore, we can conclude that \(a^3 + b^3 \neq c^3\).

It directly follows from Lemma 2.10.

Straightforward since \(n \in \mathbb {N}\) and \(\mathbb {N}\) is well-ordered.

Since \(\lambda \nmid a\), then \(\lambda ^4 \mid a^3 - 1 \lor \lambda ^4 \mid a^3 + 1\) by Lemma 2.23. Since \(\lambda \nmid b\), then \(\lambda ^4 \mid b^3 - 1 \lor \lambda ^4 \mid b^3 + 1\) by Lemma 2.23. We proceed by analysing each case:

• Case \(\lambda ^4 \mid a^3 - 1 \land \lambda ^4 \mid b^3 - 1\). Since \(\lambda \mid c\) we have that \(\lambda \mid c^3-(a^3-1)-(b^3-1) = 2\), which is absurd by Lemma 2.13.

• Case \(\lambda ^4 \mid a^3 + 1 \land \lambda ^4 \mid b^3 + 1\). Since \(\lambda \mid c\) we have that \(\lambda \mid (a^3-1)+(b^3-1)-c^3 = 2\), which is absurd by Lemma 2.13.

• Case \(\lambda ^4 \mid a^3 - 1 \land \lambda ^4 \mid b^3 + 1\). Trivial.

• Case \(\lambda ^4 \mid a^3 + 1 \land \lambda ^4 \mid b^3 - 1\). Trivial.

Apply Lemma 3.14 and then compute each case.

Apply Lemma 3.15.

It directly follows from Lemma 3.16.

It directly follows from Lemma 3.17.

Straightforward calculation using Lemma 2.16 and Lemma 2.18.

By contradiction we assume that

Then, by definition, the multiplicity of \(\lambda \) in \(a + b\), in \(a + \eta b\) and in \(a + \eta ^2 b\) is less than \(2\). By properties of divisibility, Lemma 3.16 and Lemma 3.19, we have that

Then, the multiplicity of \(\lambda \) in \((a + b) (a + \eta b) (a + \eta ^2 b)\) is greater than or equal to \(6\).
By Lemma 2.8 \(\lambda \) is prime, so we have that the multiplicity of \(\lambda \) in \((a + b) (a + \eta b) (a + \eta ^2 b)\) is the sum of the multiplicities of \(\lambda \) in \(a + b\), in \(a + \eta b\) and in \(a + \eta ^2 b\), which is less than \(6\). This is a contradiction that forces us to conclude that

By Lemma 3.20, we have that

We proceed by analysing each case:

• Case \(\lambda ^2 \mid a + b\). Trivial using \(a_1=a\) and \(b_1=b\).

• Case \(\lambda ^2 \mid a + \eta b\). Let \(a_1=a\) and \(b_1=\eta b\).
  By Lemma 2.16, we have that \(a^3 + (\eta b)^3 = a^3 + b^3 = u c^3\).
  By properties of coprimes and Lemma 2.17, we have that \(\gcd (a,b)=1\) implies that \(\gcd (a,\eta b)=1\).
  Since \(a_1=a\), we already know that \(\lambda \nmid a = a_1\).
  By contradiction we assume that \(\lambda \mid b_1 = \eta b\), which, by Lemma 2.16, it implies that \(\lambda \mid \eta ^2 \eta b = b\) that contradicts our assumption, forcing us to conclude that \(\lambda \nmid b_1\).

• Case \(\lambda ^2 \mid a + \eta ^2 b\). Let \(a_1=a\) and \(b_1=\eta ^2 b\).
  By Lemma 2.16, we have that \(a^3 + (\eta ^2 b)^3 = a^3 + b^3 = u c^3\).
  By properties of coprimes and Lemma 2.17, we have that \(\gcd (a,b)=1\) implies that \(\gcd (a,\eta ^2 b)=1\).
  Since \(a_1=a\), we already know that \(\lambda \nmid a = a_1\).
  By contradiction we assume that \(\lambda \mid b_1 = \eta ^2 b\), which, by Lemma 2.16, it implies that \(\lambda \mid \eta \eta ^2 b = b\) that contradicts our assumption, forcing us to conclude that \(\lambda \nmid b_1\).

Therefore, we can conclude that \(\exists a_1,b_1 \in \mathcal{O}_k\) such that \(S_1=(a_1,b_1,c,u)\) is a \(solution\).

Let \(S'=(a',b',c',u')\). Let \(a, b\) be the units given by Lemma 3.21. Then \(S=(a,b,c',u')\) is a \(solution'\) with multiplicity \(n\).

Trivial calculation.

Trivial since \(\lambda \mid a+b\).

Since \(\lambda \mid a+b\), then \(\lambda \mid (a + b) + \lambda ^2 b + 2 \lambda b = a + \eta ^2 b\).

By contradiction we assume that \(\lambda ^2 \mid a + \eta b\), which implies that \(\lambda ^2 \mid a + b + \lambda b\) by Lemma 3.23. Since \(\lambda ^2 \mid a+b\), then \(\lambda ^2 \mid \lambda b\), which implies that \(\lambda \mid b\), that contradicts Definition 3.8 forcing us to conclude that \(\lambda ^2 \nmid a + \eta b\).

By contradiction using Lemma 2.18, we assume \(\lambda ^2 \mid a +\eta ^2 b = a + b -b + \eta ^2 b\). Since \(\lambda ^2 \mid a+b\), then \(\lambda ^2 \mid b (\eta ^2 -1) = \lambda b (\eta + 1)\). Since \(\lambda \nmid b\), then \(\lambda \mid \eta +1 = \lambda +2\), then \(\lambda \mid 2\) which is absurd.

Trivial calculation using Lemma 2.18.

We proceed by analysis each case:

• Case \(p \mid \lambda \). It directly follows from Lemma 2.8.

• Case \(p \nmid \lambda \).
  By hypothesis, we have that \(p \mid a+b\) and \(p \mid a+\eta b\). Then \(p \mid (a+\eta b) - (a+b) = b (\eta -1) = b \lambda \), which implies that \(p \mid b\) and we proceed analogously to show that \(p \mid a\).
  Therefore \(p \mid \gcd (a,b)=1\) which is absurd.

Therefore, we can conclude that \(p\) is associated with \(\lambda \).

We proceed by analysis each case:

• Case \(p \mid \lambda \). It directly follows from Lemma 2.8.

• Case \(p \nmid \lambda \).
  By hypothesis, we have that \(p \mid a+ b\) and \(p \mid a+\eta ^2 b\). By Lemma 2.16 and Lemma 2.17, we have that

  \[ p \mid \eta ((a+\eta ^2 b) - (a+ b)) = - (\eta ^3 - \eta ) b = \lambda b, \]

  which implies that \(p \mid b\) and we proceed analogously to show that \(p \mid a\).
  Therefore \(p \mid \gcd (a,b)=1\) which is absurd.

Therefore, we can conclude that \(p\) is associated with \(\lambda \).

We proceed by analysis each case:

• Case \(p \mid \lambda \). It directly follows from Lemma 2.8.

• Case \(p \nmid \lambda \).
  By hypothesis, we have that \(p \mid a+\eta b\) and \(p \mid a+\eta ^2 b\). Then \(p \mid (a+\eta ^2 b) - (a+\eta b) = \eta b (\eta -1) = \eta b \lambda \), which, by Lemma 2.17, implies that \(p \mid b\) and we proceed analogously to show that \(p \mid a\).
  Therefore \(p \mid \gcd (a,b)=1\) which is absurd.

Therefore, we can conclude that \(p\) is associated with \(\lambda \).

By contradiction we assume that \(\lambda \mid y\), which implies that \(\lambda ^2 \mid \lambda y = a + \eta b\), that contradicts Lemma 3.26 forcing us to conclude that \(\lambda \nmid y\).

By contradiction we assume that \(\lambda \mid z\), which implies that \(\lambda ^2 \mid \lambda z = a + \eta ^2 b\), that contradicts Lemma 3.27 forcing us to conclude \(\lambda \nmid z\).

By Definition 3.10 we have that \(\lambda ^n \mid c\). Since \(u\) is a unit, then by Lemma 3.19 we have that

Then applying Lemma 3.33 and Lemma 3.34, we can conclude that \(\lambda ^{3n-2} \mid a+b\).

By contradiction we assume that \(\lambda \mid w\), which implies \(\lambda ^{n+1} \mid \lambda ^n w = c\) that contradicts Definition 3.10 forcing us to conclude that \(\lambda \nmid w\).

By contradiction, if \(\lambda \mid x\), then \(\lambda ^{3n-1} \mid \lambda ^{3n-2} x = a+b\). Using Lemma 3.24 and Lemma 3.25, we have that \(\lambda ^{3n+1} \mid (a+b) (a + \eta b) (a + \eta ^2 cdot b) = a^3+b^3 = u c^3 = u \lambda ^{3n} w^3\). Then \(\lambda \mid w^3\) which implies that \(\lambda \mid w\), that contradicts Lemma 3.36 forcing us to conclude \(\lambda \nmid x\).

Since \(y \neq 0\) by Lemma 3.33, by the properties of PIDs it suffices to prove that \(\forall p \in \mathcal{O}_K\) if \(p\) is prime and \(p \mid x\), then \(p \nmid y\). Let \(p \in \mathcal{O}_K\) be prime and suppose by contradiction that \(p \mid x\) and \(p \mid y\) which implies that \(p \mid \lambda ^{3n-2} x = a+b\) and \(p \mid \lambda y = a + \eta b\). Then by Lemma 3.29 we have that \(p\) is associated with \(\lambda \), which implies that \(\lambda \mid x\) that contradicts Lemma 3.37 forcing us to conclude that \(p \nmid y\), which, as stated above, implies that \(\gcd (x,y)=1\).

Since \(z \neq 0\) by Lemma 3.34, by the properties of PIDs it suffices to prove that \(\forall p \in \mathcal{O}_K\) if \(p\) is prime and \(p \mid x\), then \(p \nmid z\). Let \(p \in \mathcal{O}_K\) be prime and suppose by contradiction that \(p \mid x\) and \(p \mid z\) which implies that \(p \mid \lambda ^{3n-2} x = a+b\) and \(p \mid \lambda z = a + \eta ^2 b\). Then by Lemma 3.30 we have that \(p\) is associated with \(\lambda \), which implies that \(\lambda \mid x\) that contradicts Lemma 3.37 forcing us to conclude that \(p \nmid z\), which, as stated above, implies that \(\gcd (x,z)=1\).

Since \(z \neq 0\) by Lemma 3.34, by the properties of PIDs it suffices to prove that \(\forall p \in \mathcal{O}_K\) if \(p\) is prime and \(p \mid y\), then \(p \nmid z\). Let \(p \in \mathcal{O}_K\) be prime and suppose by contradiction that \(p \mid y\) and \(p \mid z\) which implies that \(p \mid \lambda y = a+\eta b\) and \(p \mid \lambda z = a + \eta ^2 b\). Then by Lemma 3.31 we have that \(p\) is associated with \(\lambda \), which implies that \(\lambda \mid y\) that contradicts Lemma 3.33 forcing us to conclude that \(p \nmid z\), which, as stated above, implies that \(\gcd (y,z)=1\).

It directly follows from Lemma 3.18 and calculations using ring properties.

It directly follows from Definition 3.32, Lemma 3.19, Lemma 2.9, Lemma 3.18 and calculations using ring properties.

By the properties of PIDs, it suffices to prove that there exists a \(k\in \mathcal{O}_K\) such that \(xk\) is a cube and \(\gcd (x,k)=1\). Let \(k = yzu^{-1}\), then \(xk = x y z u^{-1} = w^3\) by Lemma 3.42. Moreover, since \(\gcd (x,y)=1\) by Lemma 3.38 and \(\gcd (x,z)=1\) by Lemma 3.39, then \(\gcd (x,yz)=1\), which implies that \(\gcd (x,k)=1\).

By the properties of PIDs, it suffices to prove that there exists a \(k\in \mathcal{O}_K\) such that \(yk\) is a cube and \(\gcd (y,k)=1\). Let \(k = xzu^{-1}\), then \(yk = y x z u^{-1} = w^3\) by Lemma 3.42. Moreover, since \(\gcd (x,y)=1\) by Lemma 3.38 and \(\gcd (y,z)=1\) by Lemma 3.40, then \(\gcd (y,xz)=1\), which implies that \(\gcd (y,k)=1\).

By the properties of PIDs, it suffices to prove that there exists a \(k\in \mathcal{O}_K\) such that \(zk\) is a cube and \(\gcd (z,k)=1\). Let \(k = xyu^{-1}\), then \(zk = z x y u^{-1} = w^3\) by Lemma 3.42. Moreover, since \(\gcd (x,z)=1\) by Lemma 3.39 and \(\gcd (y,z)=1\) by Lemma 3.40, then \(\gcd (z,xy)=1\), which implies that \(\gcd (z,k)=1\).

By contradiction we assume that \(X = 0\), then \(x = 0\) by Definition 3.46. Therefore \(\lambda \) trivially divides \(x\) (as any number divides zero) which contradicts Lemma 3.37 forcing us to conclude that \(X \neq 0\).

By contradiction we assume that \(\lambda \mid X\), then, by the properties of divisibility, \(\lambda \mid u_1 X^3\), which implies, by Definition 3.46, that \(\lambda \mid x\). However, this contradicts Lemma 3.37 forcing us to conclude that \(\lambda \nmid X\).

By contradiction we assume that \(\lambda \mid Y\), then, by the properties of divisibility, \(\lambda \mid u_2 Y^3\), which implies, by Definition 3.46, that \(\lambda \mid y\). However, this contradicts Lemma 3.33 forcing us to conclude that \(\lambda \nmid Y\).

By contradiction we assume that \(\lambda \mid Z\), then, by the properties of divisibility, \(\lambda \mid u_3 Z^3\), which implies, by Definition 3.46, that \(\lambda \mid z\). However, this contradicts Lemma 3.34 forcing us to conclude that \(\lambda \nmid Z\).

Since \(Z \neq 0\) by Lemma 3.50, by the properties of PIDs it suffices to prove that \(\forall p \in \mathcal{O}_K\) if \(p\) is prime and \(p \mid Y\), then \(p \nmid Z\). Let \(p \in \mathcal{O}_K\) be prime and suppose by contradiction that \(p \mid Y\) and \(p \mid Z\) which implies that \(p \mid u_2 Y^3 = y\) and \(p \mid \lambda u_3 Z^3 = z\). But this contradicts Lemma 3.40 forcing us to conclude that \(p \nmid Z\), which, as stated above, implies that \(\gcd (Y,Z)=1\).

Applying Definition 3.46, Definition 3.32, Lemma 2.16 and Lemma 2.18, we have

By Definition 3.46 \(u_4 = \eta u_3 u_2^{-1}\), which is a product of units by Lemma 2.17. Since the product of units is a unit (multiplicative closure), it follows that \(u_4\) must also be a unit.

By Definition 3.46 \(u_5 = -\eta ^2 u_1 u_2^{-1}\), which is a product of units since \(\eta ^3 = 1\) by Lemma 2.16 and \(-\eta (-\eta ^2) = \eta ^3\). Since the product of units is a unit (multiplicative closure), it follows that \(u_5\) must also be a unit.

Using Lemma 2.17, Lemma 2.9, it suffices to show that

which can be proved by simple calculations involving Lemma 2.16, Lemma 3.18 and Lemma 3.52.

Straightforward application of the definition of divisibility.

Using Lemma 3.18, we have that \(\lambda ^2 \mid \lambda ^2 u_5 \lambda ^{3n-5} X^3 = u_5 (\lambda ^{n - 1} X)^3\).

Let \(n \in \mathbb {N}\) be the multiplicity of the solution \(S\).
By Theorem 2.4, it suffices to prove that

By Lemma 2.23 and Lemma 3.49, we have that

By Lemma 2.23 and Lemma 3.50, we have that

We proceed by analysing each case:

• Case \((\lambda ^4 \mid Y^3 - 1) \land (\lambda ^4 \mid Z^3 - 1)\).
  Let \(m=-1\) and consider the fact that

  \[ u_4 - m = Y^3 + u_4 Z^3 - (Y^3 - 1) - u_4 (Z^3 - 1). \]

  By Lemma 3.55, we have that

  \[ u_4 - m = u_5 (\lambda ^{n-1} X)^3 - (Y^3 - 1) - u_4 (Z^3 - 1). \]

  Since, by Lemma 3.57, we know that

  \[ \lambda ^2 \mid u_5 (\lambda ^{n-1} X)^3 \]

  and, by Lemma 3.56 and by assumption, we have that

  \[ \lambda ^2 \mid Y^3 - 1 \land \lambda ^2 \mid Z^3 - 1, \]

  Then, we can conclude that

  \[ \lambda ^2 \mid u_4 - m. \]

• Case \((\lambda ^4 \mid Y^3 - 1) \land (\lambda ^4 \mid Z^3 + 1)\).
  Let \(m=1\) and proceed similarly to the first case.

• Case \((\lambda ^4 \mid Y^3 + 1) \land (\lambda ^4 \mid Z^3 - 1)\).
  Let \(m=1\) and proceed similarly to the first case.

• Case \((\lambda ^4 \mid Y^3 + 1) \land (\lambda ^4 \mid Z^3 + 1)\).
  Let \(m=-1\) and proceed similarly to the first case.

By Lemma 3.58, we have that \(u_4 \in \left\{ -1,1\right\} \), which implies that \(u_4^2 = 1\).
Therefore, by Lemma 3.55, we can conclude that

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \((a',b',c',u') = S_f'\) be the final \(solution'\), then \(\lambda ^{n-1} \mid \lambda ^{n-1} X = c'\). By contradiction we assume that \(\lambda ^n \mid c'\) which implies that \(\lambda \mid X\), that contradicts Lemma 3.48 forcing us to conclude that \(\lambda ^{n} \nmid c'\). Then \(S_f'\) has multiplicity \(n-1\).

It directly follows from Lemma 3.61 since \(m = n-1 {\lt} n\).

It directly follows from Lemma 3.61 and Lemma 3.62.

By contradiction we assume that there are \(a, b, c \in \mathcal{O}_K\) and \(u \in \mathcal{O}^\times _K\) such that \(c \neq 0\), \(\gcd (a,b)=1\), \(\lambda \nmid a\), \(\lambda \nmid b\), \(\lambda \mid c\) and \(a^3 + b^3 = u c^3\). Then \(S'=(a,b,c,u)\) is a \(solution'\), which implies that there is a \(solution\) \(S\) by Lemma 3.22. Then, by Lemma 3.13, there is a minimal solution \(S_0\) with multiplicity \(n\). Hence, there is a \(solution'\) \(S_1'\) with multiplicity \(m{\lt}n\) by Theorem 3.63, which implies that there is a \(solution\) \(S_1\) with multiplicity \(m\) by Lemma 3.22. However, this contradicts the minimality of \(S_0\) forcing us to conclude that \(a^3 + b^3 \neq u c^3\).

Assume that \(\forall a, b, c \in \mathcal{O}_K,\, \forall u \in \mathcal{O}^\times _K\) such that \(c \neq 0\), \(\gcd (a,b)=1\), \(\lambda \nmid a\), \(\lambda \nmid b\) and \(\lambda \mid c\), it holds that \(a^3 + b^3 \neq u c^3\). Let \(a, b, c \in \mathbb {Z}\) such that \(a\neq 0\), \(b\neq 0\) and \(c\neq 0\). By Theorem 3.6, we can assume that \(\gcd (a,b)=1\), \(3 \nmid a\), \(3 \nmid b\), \(3 \mid c\). By contradiction we assume that \(a^3 + b^3 = c^3\) and let \(u = 1\).

• By contradiction we assume that \(\lambda \mid a\), which implies that the norm of \(\lambda \) divides \(a\) by Lemma 2.6, which implies that \(3 \mid a\) by Lemma 2.5, that contradicts the assumption that \(3 \nmid a\) forcing us to conclude that \(\lambda \nmid a\).

• By contradiction we assume that \(\lambda \mid b\), which implies that the norm of \(\lambda \) divides \(b\) by Lemma 2.6, which implies that \(3 \mid b\) by Lemma 2.5, that contradicts the assumption that \(3 \nmid b\) forcing us to conclude that \(\lambda \nmid b\).

• \(\lambda \mid 3\) by Lemma 2.7 and \(3 \mid c\), then \(\lambda \mid c\).

By our first assumption \(a^3 + b^3 \neq u c^3 = 1 c^3 = c^3 = a^3 + b^3\) which is absurd.

By Lemma 3.65 and Theorem 3.64, we can conclude that