There are no solutions of cubic Fermat equation in

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Nov 28, 2018 - Annals of. Mathematics 142 (3): 443–551. [2] Bourbaki, Nicolas (1994). Elements of the history of mathematics. Translated by Meldrum,. John.
There are no solutions of cubic Fermat equation in Gaussian integers Mercedes Or´ us–Lacort Online teacher of College Mathematics November 28, 2018

Abstract In this paper we show that there are no 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 integers field, and they specified how we could write those quadratic integers solution. [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 no solution in Gaussian integers. This article is organized as follows: in Sec 2 we show that there are no solutions of Fermat cubic equation in Gaussian integers, and in Sec 3 we wrap up our conclusions.

1

2

There are no 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 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 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

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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. Therefore there are not solutions in Gaussian integers for cubic Fermat equation.

3

Conclusions

We have shown how to prove that the cubic Fermat equation has no 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 no solution in Gaussian integers. Acknowledgements: we acknowledge Roman Orus for proposing the dissemination of this result.

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[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]

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