There are not non trivial solutions of cubic Fermat

Download PDF
0 downloads 0 Views 214KB Size Report
Comment
Dec 13, 2018 - It is well known from Fermat's last theorem, proved by Andrew Wiles [1], ... Proof: If 2a(a2 + 3b2D) = e3, then e is a multiple of 2, that is, there ...
There are not non trivial solutions of cubic Fermat equation in Gaussian integers Mercedes Or´ us–Lacort Online teacher of College Mathematics December 13, 2018

Abstract In this paper we show that there are not non trivial solutions of Fermat cubic equation in Gaussian integers, as a consequence of how the solutions of the Fermat cubic equation are in quadratic integers.

Key Words: Number theory, Cubic Fermat equation, Gaussian integers, Quadratic integers. AMS Classification: 11–02

1

Introduction

It is well known from Fermat’s last theorem, proved by Andrew Wiles [1], that the equation xn + y n = z n has no solutions x, y, z in Z for n > 2. This circumstance has led mathematicians to study extensions of this equation in other sets of numbers, such as quadratic integers [2] [3] [4] [5]. Concerning the quadratic integers it has been proved that the equation x3 + y 3 = z 3 has solution. Mathematicians as Rudolf Fueter, William Burnside, Francisco Jos´e Duarte, T. Nagell, E. Fogels, and Alexander Aigner were the first to demonstrate that the cubic Fermat equation had solutions in the quadratic field, and they specified how we could write those solutions. [6] [7] [8] [9] [10] [11] [12]. Thanks to them, √ it is well known √ that the quadratic integers solution of said equation can be written as: x = a + b D, y = a − b D, z = e, for {a, b, c, d, e, D} ⊂ Z \ {0}, and |D| not a square. √ √ In this article, we show that from x = a + b D, y = a − b D, z = e satisfying the equation x3 + y 3 = z 3 , we can deduce a solution form for quadratic integers, and the corresponding D to these solutions, can not be neither equal to −1 nor to −n2 for n ∈ Z, and therefore, the cubic Fermat equation has not non trivial solution in Gaussian integers. This article is organized as follows: in Sec 2 we show that there are not non trivial solutions of Fermat cubic equation in Gaussian integers, and in Sec 3 we wrap up our conclusions.

1

2

There are not non trivial solutions of Fermat cubic equation in Gaussian integers.

√ √ If x = a + b D, y = a − b D, z = e, for {a, b, c, d, e, D} ⊂ Z \ {0}, and |D| not a square, is a solution of x3 + y 3 = z 3 , then the condition 2a(a2 + 3b2 D) = e3 is satisfied. We remember this condition because it will be used in the development of this demonstration. Lemma: If 2a(a2 + 3b2 D) = e3 , then for any k ∈ Z \ {0}, the solutions of Fermat cubic equation in quadratic integers can be written as:

p −3a(a3 − 4k 3 ) p Y = 3a2 − −3a(a3 − 4k 3 )

X = 3a2 +

(1)

Z = 6ka Proof: If 2a(a2 + 3b2 D) = e3 , then e is a multiple of 2, that is, there exists k ∈ Z \ {0}, such that e = 2k. This implies:

2a(a2 + 3b2 D) = 8k 3 a(a2 + 3b2 D) = 4k 3

(2)

4k 3 − a3 D= 3ab2

Therefore,

√

1 D= b

r

√ (4k 3 − a3 ) , and the solutions (x, y, z) in Q( D) are: 3a r r 1 4k 3 − a3 4k 3 − a3 x=a+b· =a+ br 3a 3a r 3 3 3 1 4k − a 4k − a3 y =a−b· =a− b 3a 3a z = 2k

(3)

√ Now, multiplying by 3a, we obtain the equivalent solutions (X, Y, Z) in Z( D), stated in the Lemma. That is:

X = 3a2 +

p

−3a(a3 − 4k 3 ) p Y = 3a2 − −3a(a3 − 4k 3 )

(4)

Z = 6ka Theorem: −3a(a3 − 4k 3 ) 6= −n2 , for n ∈ Z \ {0}, and therefore, there are not non trivial solutions for cubic Fermat equation in Gaussian integers. 2

Proof: p If −3a(a3 − 4k 3 ) = −n2 for any a, k and n in Z \ {0}, then −3a(a3 − 4k 3 ) = i · n, and therefore, there would be non trivial solutions in Gaussian integers for cubic Fermat equation. We will prove that −3a(a3 − 4k 3 ) 6= −n2 for any a, k and n in Z \ {0}. First: 3a(a3 − 4k 3 ) 6= 1 Since a ∈ Z \ {0}, this implies 3a ≥ 3 or 3a ≤ −3. On the other hand, since k ∈ Z \ {0}, a3 − 4k 3 will be an integer number, greater than 1 or less than −1. Therefore, 3a(a3 − 4k 3 ) will be greater than 3 or less than −3, but never equal to 1. Second: 3a(a3 − 4k 3 ) 6= n2 Let us assume that 3a(a3 − 4k 3 ) = n2 . If 3a(a3 − 4k 3 ) = n2 implies n2 = 3r, for r ∈ Z \ {0}, that is, n = 3m for m ∈ Z \ {0}. Hence, 3a(a3 − 4k 3 ) = 32 m2 , that is, a(a3 − 4k 3 ) = 3m2 . Since m ∈ Z \ {0}, let us assume that m = pn1 1 · pn2 2 · ... · pns s , for pi ∈ Z \ {0} and ni ∈ N \ {0}, i = 1, 2, ..., s. Hence, we will call P to any product of the factors of m (P different of m and −m), and we will call Q to the result of dividing m by P . We will analyze if there exist a, k and m in Z \ {0} satisfying a(a3 − 4k 3 ) = 3m2 . To analyze it, we will solve the equation a4 − 4ak 3 − 3m2 = 0 for a, and we will analyze if the values obtained for a and k are in Z \ {0} or not. Let P (a) = a4 − 4ak 3 − 3m2 . Using the polynomial remainder theorem, and the rational root theorem, the possible a roots values in Z\{0} are the possible integer divisors of 3m2 , that is: {1, −1, 3, −3, m, −m, 3m, −3m, m2 , − m2 , 3m2 , −3m2 , P, −P, 3P. − 3P }. Hence, we have: r

1 − 3m2 , however this k 6∈ Z, because if m 4 2 is even, that is, m = 2t for t ∈ Z \ {0}, then 1 − 3m = 1 − 12t2 ≡ 1 (mod 4), and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then 1 − 3m2 = 1 − 12t2 − 12t − 3 ≡ −2 ≡ 2 (mod 4). Hence, 1 − 3m2 is not a multiple of 4, that is, 1 − 3m2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 2 3 3m − 1 • If a = −1 then 1 + 4k 3 − 3m2 = 0, that is, k = , however this k 6∈ Z, because, if 4 m is even, that is, m = 2t for t ∈ Z \ {0}, then 3m2 − 1 = 12t2 − 1 ≡ −1 ≡ 3 (mod 4), and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then 3m2 − 1 = 12t2 + 12t + 3 − 1 ≡ 2 (mod 4). Hence, 3m2 − 1 is not a multiple of 4, that is 3m2 − 1 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r r 2 2 81 − 3m 3 3 27 − m • If a = 3 then 81 − 12k 3 − 3m2 = 0, that is, k = = , however this 12 4 k 6∈ Z, because, if m is even, that is, m = 2t for t ∈ Z \ {0}, then 27 − m2 = 27 − 4t2 ≡ 3 (mod 3

2

• If a = 1 then 1 − 4k − 3m = 0, that is, k =

3

3

•

•

•

•

•

•

•

4), and if m is odd, that is m = 2t+1 for t ∈ Z\{0}, then 27−m2 = 27−4t2 −4t−1 ≡ 26 ≡ 2 (mod 4). Hence, 27−m2 is not a multiple of 4, that is 27−m2 can not be u3 ·4 for u ∈ Z\{0}. Therefore, this case is not possible. r r 2 2 3 m − 27 3 3m − 81 3 2 = , however this If a = −3 then 81 + 12k − 3m = 0, that is, k = 12 4 2 2 k 6∈ Z, because, if m is even, that is, m = 2t for t ∈ Z \ {0}, then m − 27 = 4t − 27 ≡ 3 (mod 4), and if m is odd, that is m = 2t+1 for t ∈ Z\{0}, then m2 −27 = 4t2 +4t+1−27 ≡ −26 ≡ 2 (mod 4). Hence, m2 −27 is not a multiple of 4, that is m2 −27 can not be u3 ·4 for u ∈ Z\{0}. Therefore, this case is not possible. r 3 3 m − 3m 4 3 2 If a = m then m −4mk −3m = 0, that is, k = , however this k 6∈ Z, because, if 4 3 3 m is even, that is, m = 2t for t ∈ Z \ {0}, then m − 3m = 8t − 6t ≡ 0 (mod 4) only if t = 2r 64r3 − 12r = 16r3 − 3r = u3 only has integer solutions if r = u = 0, for r ∈ Z \ {0}, however 4 and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then m3 − 3m = 8t3 + 12t2 − 2 ≡ 2 (mod 4). Hence, m3 − 3m is not a multiple of 4, that is m3 − 3m can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 3 3 3m − m 4 3 2 If a = −m then m + 4mk − 3m = 0, that is, k = , however this k 6∈ Z, because, 4 3 if m is even, that is, m = 2t for t ∈ Z\{0}, then 3m−m = 6t−8t3 ≡ 0 (mod 4) only if t = 2r 12r − 64r3 for r ∈ Z \ {0}, however = 3r − 16r3 = u3 only has integer solutions if r = u = 0, 4 and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then 3m − m3 = 2 − 8t3 − 12t2 ≡ 2 (mod 4). Hence, 3m − m3 is not a multiple of 4, that is 3m − m3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 3 3 27m − m 4 3 2 If a = 3m then 81m − 12mk − 3m = 0, that is, k = , however this k 6∈ Z, 4 because, if m is even, that is, m = 2t for t ∈ Z \ {0}, then 27m3 − m = 216t3 − 2t ≡ 0 1728r3 − 4r (mod 4) only if t = 2r for r ∈ Z \ {0}, however = 432r3 − r = u3 only has 4 integer solutions if r = u = 0, and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then 27m3 − m = 216t3 + 324t2 + 160t + 26 ≡ 2 (mod 4). Hence, 27m3 − m is not a multiple of 4, that is 27m3 − m can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 3 3 m − 27m 4 3 2 If a = −3m then 81m + 12mk − 3m = 0, that is, k = , however this k 6∈ Z, 4 because, if m is even, that is, m = 2t for t ∈ Z \ {0}, then m − 27m3 = 2t − 216t3 ≡ 0 4r − 1728r3 (mod 4) only if t = 2r for r ∈ Z \ {0}, however = r − 432r3 = u3 only has 4 integer solutions if r = u = 0, and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then m − 27m3 = −216t3 − 324t2 − 160t − 26 ≡ 2 (mod 4). Hence, m − 27m3 is not a multiple of 4, that is m − 27m3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 6 3 m − 3 2 8 2 3 2 , however this k 6∈ Z, because, if If a = m then m −4m k −3m = 0, that is, k = 4 6 6 m is even, that is, m = 2t for t ∈ Z\{0}, then m −3 = 64t −3 ≡ 1 (mod 4), and if m is odd, that is m = 2t+1 for t ∈ Z\{0}, then m6 −3 = 64t6 +192t5 +240t4 +160t3 +60t2 +12t−2 ≡ 2 (mod 4). Hence, m6 − 3 is not a multiple of 4, that is m6 − 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 6 3 3 − m If a = −m2 then m8 + 4m2 k 3 − 3m2 = 0, that is, k = , however this k 6∈ Z, because, 4 if m is even, that is, m = 2t for t ∈ Z\{0}, then 3−m6 = 3−64t6 ≡ 3 (mod 4), and if m is odd, 4

that is m = 2t+1 for t ∈ Z\{0}, then 3−m6 = −64t6 −192t5 −240t4 −160t3 −60t2 −12t+2 ≡ 2 (mod 4). Hence, 3 − m6 is not a multiple of 4, that is 3 − m6 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 6 3 27m − 1 2 8 2 3 2 • If a = 3m then 81m − 12m k − 3m = 0, that is, k = , however this k 6∈ Z, 4 6 because, if m is even, that is, m = 2t for t ∈ Z \ {0}, then 27m − 1 = 1728t6 − 1 ≡ 3 (mod 4), and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then 27m6 − 1 = 1728t6 + 5184t5 + 6480t4 + 4320t3 + 1620t2 + 324t + 26 ≡ 2 (mod 4). Hence, 27m6 − 1 is not a multiple of 4, that is 27m6 − 1 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. r 6 3 1 − 27m 2 8 2 3 2 , however this k 6∈ Z, • If a = −3m then 81m + 12m k − 3m = 0, that is, k = 4 because, if m is even, that is, m = 2t for t ∈ Z \ {0}, then 1 − 27m6 = 1 − 1728t6 ≡ 1 (mod 4), and if m is odd, that is m = 2t + 1 for t ∈ Z \ {0}, then 1 − 27m6 = −1728t6 − 5184t5 − 6480t4 − 4320t3 − 1620t2 − 324t − 26 ≡ 2 (mod 4). Hence, 1 − 27m6 is not a multiple of 4, that is 1 − 27m6 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. 4 3 2 4 3 2 2 • If a = rP then P − 4P k − 3m = 0, or writing m as P Q, P − 4P k − 3P Q = 0, that is 3 2 − 3P Q 3 P , however this k 6∈ Z, because: k= 4

– if P and Q are even, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w for w ∈ Z \ {0}, P 3 − 3P Q2 then = 2v 3 − 6vw2 , however 2v 3 − 6vw2 = u3 only has integer solutions if 4 v = u = 0. Therefore, this case is not possible. – if P and Q are odd, that is, P = 2v + 1 for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then P 3 − 3P Q2 = 8v 3 + 12v 2 − 24vw2 − 24vw − 12w2 − 12w − 2 ≡ 2 (mod 4). Hence, P 3 − 3P Q2 is not a multiple of 4, that is P 3 − 3P Q2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is even and Q is odd, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then P 3 − 3P Q2 = 8v 3 − 24vw2 − 24vw − 6v ≡ 0 (mod 4) only if v = 2r P 3 − 3P Q2 64r3 − 48rw2 − 48rw − 12r = = for r ∈ Z \ {0}, however in this case 4 4 3 2 3 2 3 16r − 12rw − 12rw − 3r, and 16r − 12rw − 12rw − 3r = u only has integer solutions if r = u = 0. Hence, P 3 − 3P Q2 is not a multiple of 4, that is P 3 − 3P Q2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is odd and Q is even, that is, P = 2v +1 for v ∈ Z\{0}, and Q = 2w for w ∈ Z\{0}, then P 3 − 3P Q2 = 8v 3 + 12v 2 − 24vw2 + 6v − 12w2 + 1 ≡ 1 (mod 4). Hence, P 3 − 3P Q2 is not a multiple of 4, that is P 3 − 3P Q2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. 4 3 2 4 3 2 2 • If a = −P r then P + 4P k − 3m = 0, or writing m as P Q, P + 4P k − 3P Q = 0, that 2 3 3 3P Q − P is k = , however this k 6∈ Z, because: 4

– if P and Q are even, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w for w ∈ Z \ {0}, then 3P Q2 − P 3 = −2v 3 + 6vw2 , however −2v 3 + 6vw2 = u3 only has integer solutions if 4 v = u = 0. Therefore, this case is not possible. – if P and Q are odd, that is, P = 2v + 1 for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then 3P Q2 − P 3 = −8v 3 − 12v 2 + 24vw2 + 24vw + 12w2 + 12w + 2 ≡ 2 (mod 4). Hence, 3P Q2 − P 3 is not a multiple of 4, that is 3P Q2 − P 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. 5

– if P is even and Q is odd, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then 3P Q2 − P 3 = −8v 3 + 24vw2 + 24vw + 6v ≡ 0 (mod 4) only if v = 2r 3P Q2 − P 3 −64r3 + 48rw2 + 48rw + 12r for r ∈ Z \ {0}, however in this case = = 4 4 3 2 3 2 3 −16r + 12rw + 12rw + 3r, and −16r + 12rw + 12rw + 3r = u only has integer solutions if r = u = 0. Hence, 3P Q2 − P 3 is not a multiple of 4, that is 3P Q2 − P 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is odd and Q is even, that is, P = 2v +1 for v ∈ Z\{0}, and Q = 2w for w ∈ Z\{0}, then 3P Q2 −P 3 = −8v 3 −12v 2 +24vw2 −6v +12w2 −1 ≡ 3 (mod 4). Hence, 3P Q2 −P 3 is not a multiple of 4, that is 3P Q2 − P 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. 4 3 2 4 3 2 2 • If a = 3P then r 81P − 12P k − 3m = 0, or writing m as P Q, 81P − 12P k − 3P Q = 0, 3 − P Q2 3 27P , however this k 6∈ Z, because: that is k = 4

– if P and Q are even, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w for w ∈ Z \ {0}, then 27P 3 − P Q2 = 54v 3 − 2vw2 , however 54v 3 − 2vw2 = u3 only has integer solutions if 4 v = u = 0. Therefore, this case is not possible. – if P and Q are odd, that is, P = 2v + 1 for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then 27P 3 − P Q2 = 216v 3 + 324v 2 − 8vw2 − 8vw + 160v − 4w2 − 4w + 26 ≡ 2 (mod 4). Hence, 27P 3 − P Q2 is not a multiple of 4, that is 27P 3 − P Q2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is even and Q is odd, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then 27P 3 − P Q2 = 216v 3 − 8vw2 − 8vw − 2v ≡ 0 (mod 4) only if v = 2r 1728r3 − 16rw2 − 16rw − 4r 27P 3 − P Q2 = = for r ∈ Z \ {0}, however in this case 4 4 3 2 3 2 3 432r − 4rw − 4rw − r, and 432r − 4rw − 4rw − r = u only has integer solutions if r = u = 0. Hence, 27P 3 − P Q2 is not a multiple of 4, that is 27P 3 − P Q2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is odd and Q is even, that is, P = 2v +1 for v ∈ Z\{0}, and Q = 2w for w ∈ Z\{0}, then 27P 3 − P Q2 = 216v 3 + 324v 2 − 8vw2 + 162v − 4w2 + 27 ≡ 3 (mod 4). Hence, 27P 3 − P Q2 is not a multiple of 4, that is 27P 3 − P Q2 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. • If a = −3P r then 81P 4 + 12P k 3 − 3m2 = 0, or writing m as P Q, 81P 4 + 12P k 3 − 3P 2 Q2 = 0, 2 3 3 P Q − 27P that is k = , however this k 6∈ Z, because: 4 – if P and Q are even, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w for w ∈ Z \ {0}, then P Q2 − 27P 3 = −54v 3 + 2vw2 , however −54v 3 + 2vw2 = u3 only has integer solutions 4 if v = u = 0. Therefore, this case is not possible. – if P and Q are odd, that is, P = 2v + 1 for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then P Q2 − 27P 3 = −216v 3 − 324v 2 + 8vw2 + 8vw − 160v + 4w2 + 4w − 26 ≡ 2 (mod 4). Hence, P Q2 − 27P 3 is not a multiple of 4, that is P Q2 − 27P 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is even and Q is odd, that is, P = 2v for v ∈ Z \ {0}, and Q = 2w + 1 for w ∈ Z \ {0}, then P Q2 − 27P 3 = −216v 3 + 8vw2 + 8vw + 2v ≡ 0 (mod 4) only if v = 2r −1728r3 + 16rw2 + 16rw + 4r P Q2 − 27P 3 = = for r ∈ Z \ {0}, however in this case 4 4 3 2 3 2 3 −432r + 4rw + 4rw + r, and −432r + 4rw + 4rw + r = u only has integer solutions

6

if r = u = 0. Hence, P Q2 − 27P 3 is not a multiple of 4, that is P Q2 − 27P 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. – if P is odd and Q is even, that is, P = 2v +1 for v ∈ Z\{0}, and Q = 2w for w ∈ Z\{0}, then P Q2 − 27P 3 = −216v 3 − 324v 2 + 8vw2 − 162v + 4w2 − 27 ≡ 1 (mod 4). Hence, P Q2 − 27P 3 is not a multiple of 4, that is P Q2 − 27P 3 can not be u3 · 4 for u ∈ Z \ {0}. Therefore, this case is not possible. Hence, −3a(a3 − 4k 3 ) 6= −n2 , for n ∈ Z \ {0}, and therefore, there are not non trivial solutions for cubic Fermat equation in Gaussian integers. The only possible solutions in Gaussian integers are the trivial solutions, (x, −x, 0), or (−y, y, 0), or (x, 0, x), or (0, y, y) for x ∈ Z[i] .

3

Conclusions

We have shown how to prove that the cubic Fermat equation has not non trivial solution in Gaussian integers. √ √ We show that from x = a + b D, y = a − b D, z = e satisfying the equation x3 + y 3 = z 3 , we can deduce a solution form for quadratic integers, and the corresponding D to these solutions, can not be neither equal to −1 nor to −n2 for n ∈ Z, and therefore, the cubic Fermat equation has not non trivial solution in Gaussian integers. Acknowledgements: we acknowledge Roman Orus and Jean-Claude Evard for proposing the dissemination of this result.

References [1] Andrew Wiles (1995). Modular elliptic curves and Fermat’s Last Theorem. Annals of Mathematics 142 (3): 443–551. [2] Bourbaki, Nicolas (1994). Elements of the history of mathematics. Translated by Meldrum, John. Berlin: Springer-Verlag. ISBN 978-3-540-64767-6. MR 1290116. [3] Dedekind, Richard (1871), Vorlesungen u ¨ber Zahlentheorie von P.G. Lejeune Dirichlet (2 ed.), Vieweg. Retrieved 5. August 2009 [4] Dummit, D. S., and Foote, R. M., 2004. Abstract Algebra, 3rd ed. [5] Artin, M, Algebra, 2nd ed., Ch 13. [6] Fueter, Rudolf (1913). Die diophantische Gleichung ξ 3 + ρ3 + ζ 3 = 0. Sitzungsberichte der Heidelberger Akademie der Wissenschaften Mathematisch-Naturwissenschaftliche Klasse : A Mathematisch-physikalische Wissenschaften, 1913, 25 December 19, 1913 [7] Burnside, William (1915). On the Rational Solutions of the Equation X 3 + Y 3 + Z 3 = 0 In Quadratic Fields. Proceedings of the London Mathematical Society Volume s2 14, issue 1, page 1—4. [8] Duarte, F. J. (1933). Sur l’Equation x3 + y 3 = z 3 . L’Enseignement Math´ematique, 32, pages 68—74, (1933). DOI: 10.5169/seals-25321 [9] Nagell, T. (1936). Bemerkungen u ¨ber die Diophantische Gleichung x3 + y 3 = Az 3 . Arkiv Mat. Astr. Fysik, 25B (1936), no. 5, 6 pp. 7

¨ [10] Fogels, E. (1937). Uber die M¨ oglichkeit einiger diophantischer Gleichungen 3. und 4Grades in quadratischen K¨ orpern Commentarii. Mathematici Helvetici, 10/de (1937-1938) [11] Aigner, Alexander. (1952). Weitere Ergebnisse u ¨ber x3 + y 3 = z 3 in quadratischen K¨orpern. Monatshefte f¨ ur Mathematik Volume 56, Issue 3, pp 240–252. DOI: 10.1007/BF01297498, September (1952) [12] Aigner, Alexander. Die kubische Fermatgleichung in quadratischen K¨orpern. Journal f¨ ur die reine und angewandte Mathematik (Crelle’s Journal), Volume 1955, issue 19 DOI: 10.1515/crll.1955.195.3 January 1, (1955) Email: [email protected]

8