Solutions of cubic Fermat equation in quadratic integers in non conjugate form Mercedes Or´ us–Lacort Online teacher of College Mathematics November 30, 2018
Abstract In this paper we show that there are solutions of the Fermat cubic equation, in quadratic integers, in non conjugated form. We use the method developed by Alexander Aigner in 1952, where he improved the methods of William Burnside and Francisco Jos´e Duarte, concerning to how to find the solutions form, in quadratic fields, for the cubic Fermat equation. We show two possible algorithms to convert one conjugate solution form into another non conjugated solution form.
Key Words: Number theory, Cubic Fermat equation, Gaussian integers, Quadratic integers. AMS Classification: 11–02
1
Introduction
Alexander Aigner in 1952 [3], improved the methods of William Burnside [1], and Francisco Jos´e Duarte [1], concerning to how to find the solutions form, in quadratic fields, for the cubic Fermat equation. √ √ The method of Alexander Aigner begins by assuming that there are a1 + b1 D, a2 + b2 D and e, for {a1 , b1 , a2 , b2 , e} ⊂ Q \ {0} satisfying the cubic equation. That is: √ √ (a1 + b1 D)3 + (a2 + b2 D)3 = e3
(1)
Using the above equality, he obtained the possible values of a1 , a2 and D. And these values were:
(−b31 + 2b32 )e 2(b31 + b32 ) −3b21 · b2 · e a2 = 2(b31 + b32 ) −3b1 (b31 + 4b32 )e2 D= 4(b31 + b32 )2 a1 =
1
(2)
Therefore, with these a1 , a2 and D values he wrote a first solution of the cubic Fermat equation as:
q x = −b31 + 2b32 + b1 −3b1 (b31 + 4b32 q y = −3b21 · b2 + b2 −3b1 (b31 + 4b32 )
(3)
z = 2(b31 + b32 ) Notice that, in the above solution form, y was not the conjugate of x. Aigner used Eq. 3 to obtain another solution form, where y was x conjugate. This solution form was:
X = 3b21 +
q
−3b1 (b31 + 4b32 ) q Y = 3b21 − −3b1 (b31 + 4b32 )
(4)
Z = −6b1 b2 In this paper we will use both solutions forms (Eq 3 and 4), to show two algorithms to convert a conjugate solution into a non conjugate solution. This article is organized as follows: in Sec 2 we show how to convert a conjugate solution of the cubic Fermat equation into a non conjugate solution, and in Sec 3 we wrap up our conclusions.
2
How to convert a conjugate solution of the cubic Fermat equation into a non conjugate solution.
√ √ Let us assume that x = a + b D, y = a − b D and z = e, is a solution of cubic Fermat equation. We will show two algorithms to convert a conjugate solution. These algorithms are based on the two form solutions obtained by Alexander Aigner Eq. 3 and 4, and also in how it is the solution that we want to convert. More specifically, the use of one or another algorithm will depend on how are the values of a and b. First algorithm: When a = 3n2 and b = 1 √ √ Let us assume that our solution x = a + b D, y = a − b D and z = e, is such that a = 3n2 for n ∈ Z \ {0} and b = 1. If we use the solution form (X, Y, Z) from Eq. 4, that is:
X = 3b21 +
q
−3b1 (b31 + 4b32 ) q Y = 3b21 − −3b1 (b31 + 4b32 ) Z = −6b1 b2 2
(5)
We can calculate the values of b1 and b2 , using the formulas 3b21 = 3n2 and −3b1 (b31 + 4b32 ) = D, that is:
3b21 = 3n2
(6)
b1 = ±n And now, with b1 we can obtain b2 , that is: r
−D − 3n4 . 12n r 4 3 D + 3n 3 3 3 3 . • If b1 = −n, then −3b1 (b1 + 4b2 ) = 3n(−n + 4b2 ) = D, that is, b2 = 12n • If b1 = n, then
−3b1 (b31
+
4b32 )
3
= −3n(n +
4b32 )
= D, that is, b2 =
3
Therefore, with b1 and b2 values, we can write the non conjugate solution using the Aigner formulas from Eq. 3:
q x = −b31 + 2b32 + b1 −3b1 (b31 + 4b32 q y = −3b21 ∗ b2 + b2 −3b1 (b31 + 4b32 )
(7)
z = 2(b31 + b32 ) Let us see an example: √ √ √ Let X = 3 + −15, √ Y = 3 − −15, Z = −6, be a solution, where as we can see 3 − −15 is the conjugate of 3 + −15. If we want to obtain a new solution (x, y, z) where y will not be the conjugate of x, we must follow these steps: First step: Using the formulas of X, Y and Z (Eq. 4), and using our X = 3 +
√
−15, we have:
3b21 = 3 b21 = 1
(8)
b1 = ±1 Second step: We will consider b1 = 1, and we calculate b2 as follows: − 3b1 (b31 + 4b32 ) = −15 − 3(1 + 4b32 ) = −15 1 + 4b32 = 5 4b32
=4
b2 = 1 3
(9)
Third step: Therefore, if b1 = 1 and b2 = 1, using the formulas for x, y and z (Eq. 3), we have this another solution:
q p √ x = −b31 + 2b32 + b1 −3b1 (b31 + 4b32 ) = −1 + 2 + −3(1 + 4) = 1 + −15 q p √ y = −3b21 · b2 + b2 −3b1 (b31 + 4b32 ) = −3 + −3(1 + 4) = −3 + −15
(10)
z = 2(b31 + b32 ) = 2(1 + 1) = 4 √ √ That is, from the solution √ X = 3 + −15, √ Y = 3 − −15, and Z = −6, we have a new non conjugate solution x = 1 + −15, y = −3 + −15, and z = 4. Second algorithm: When a 6= 3n2 and/or b 6= 1 Let us assume that our solution is (x, y, z), where a 6= 3n2 and/or b 6= 1. When a 6= 3n2 and/or b 6= 1, we can not use the form solution Eq. 4 to obtain b1 and b2 . Hence, in this case we must follow these steps: First step: We multiply our solution by
√
√ √ √ D, and we obtain (x D, y D, z D).
Second step: √ √ √ We write the following equivalent solution (x D, −z D, −y D) and we simplify it if we need it. Third step: With this solution form, we can use the formulas for a1 , a2 and D from Eq. 2, and we can write a new solution in non conjugate form as it is shown in Eq. 3. Let us see an example. √ √ √ Let X = 4 + √ −5, Y = 4 − −5, Z = 2, be a solution, where as we can see 4 − −5 is the conjugate of 4 + −5. If we want to obtain a new solution (x, y, z) where y will not be the conjugate of x, we must follow these steps: First step: We multiply our solution by
√
√ √ √ −5, and we obtain (−5 + 4 −5, 5 + 4 −5, 2 −5).
Second step: √ √ √ We write the following equivalent solution (−5 + 4 −5, −2 −5, −5 − 4 −5) and we simplify it if we need it. Third step: With this solution form, and b1 = 4 and b2 = −2, we can write a new solution in non conjugate form as it is shown in Eq. 3. That is:
4
q √ x = −b31 + 2b32 + b1 −3b1 (b31 + 4b32 = −80 + 4 384 q √ y = −3b21 · b2 + b2 −3b1 (b31 + 4b32 ) = 96 − 2 384
(11)
z = 2(b31 + b32 ) = 112 √ Hence, √ we have obtained a new solution in non conjugate form, x = −80 + 4 384, y = this solution can be simplified, and simplifying this solution 96 − 2 384 and z = 112. However, √ √ we finally obtain x = −10 + 2 2, y = 12 − 2 and z = 14, a non conjugate solution completely simplified. The following tables shows the equivalent solutions converted into non conjugate form, from a set of solutions given in conjugate form. Table 1 In Conjugate Form a
b
c
d
e
In No Conjugate Form Simplified (with f=0) f
D
a
b
c
d
e
D
17
-9 0
21
17
9
2
713
-9
-189
21
632
1345
5
-9 0
12
5
9
5
155
-9
-108
12
74
229
5
-1 0
3
5
1
6
55
-1
-9
3
52
321
215
-4 0
42
215
4 15
9265
-2
-126
21
9253
55554
2159
-36 0
390
2159
36 17
274733
-18
-3510
195
274409
549142
11
-4 0
6
11
4 33
31
-10
-18
15
19
6
473
-4 0
50
473
4 43
15629
-2
-150
25
15617
93738
8207
-8 0
382
8207
8 58
6967903
-8
-4584
382
6967807
20903565
1265
-27 0
156
1265
27 69
281945
-27
-12636
156
279758
1685109
41
-1 0
5
41
1 82
251
-1
-15
5
248
1497
85
-1 0
8
85
1 85
1025
-1
-24
8
1022
6141
31
-1 0
4
31
1 93
43
-1
-4
4
42
85
1
-8 0
10
1
8
-2
157
-8
-120
10
61
327
5
-4 0
2
5
4
-5
5
-2
-6
1
-7
-6
1
-1 0
1
1
1
-6
1
-1
-1
1
0
0
1705
-972 0
666
1705
972 -11
129385
-486
-161838
333
-106811
22574
11
-5 0
2
11
5 -15
47
-5
-50
2
-78
-155
-23
-9 0
6
-23
9 -23
43
-9
-54
6
-38
5
307
-117 0
87
307
117 -26
108097
-117
-132327
87
-69860
497081
31
-243 0
306
31
243 -31
98291
-243
-74358
306
39242
137533
451
-256 0
296
451
256 -41
67037
-128
-56832
148
17885
254766
5953 -1504 0
20
5953 1504 -47
26578813
-752
-1060320
10
-53157251 -3747559758
1727
-544 0
508
1727
544 -51
1102037
-272
-1174496
254
-155691
-16087882
1559
-531 0
552
1559
531 -59
18004241
-531
-17293608
552
1368542 1142994197
14279 -2752 0 1364 14279 2752 -86 202482157 -2752 -242115456 1364 -286008851 -5387471763 145
-27 0
6
145
27 -87
745
-27
-486
6
-1442
-2091
89
-36 0
42
89
36 -89
451
-306
-378
357
127
2
5
a -5 109 11 -227 -307 -899 -287 -25.451 -6.443 -47 727 -449 2.281 -1.643 473 -4.061
3
Table 2 In Conjugate Form In No Conjugate Form b c d e f D a equiv b equiv c equiv d equiv e equiv 4 0 -2 -5 -4 -5 -10 2 12 -1 14 4 0 -14 109 -4 109 -2776 2 336 -7 -2680 4 0 -6 11 -4 33 -248 2 144 -3 -152 32 0 -28 -227 -32 -227 -19168 104 21504 -91 5408 36 0 -30 -307 -36 -307 -16776 6 19440 -5 6552 32 0 -20 -899 -32 -2697 -4064 8 5120 -5 4128 18 0 -11 -287 -18 -861 -8494 18 10692 -11 9002 288 0 -156 -25.451 -288 -25451 -2623392 24 3234816 -13 3348576 144 0 -74 -6.443 -144 -6443 -1898216 72 2301696 -37 2580760 4 0 10 -47 -4 -47 968 2 -240 5 1064 4 0 -26 727 -4 727 -17608 2 624 -13 -17512 4 0 22 -449 -4 -449 10616 2 -528 11 10712 4 0 -38 2.281 -4 2281 -54904 2 912 -19 -54808 4 0 34 -1.643 -4 -1643 39272 2 -816 17 39368 4 0 -50 473 -4 43 -125032 2 1200 -25 -124936 4 0 46 -4.061 -4 -4061 97304 10 -1104 115 97400
D equiv 2 -670 -38 2 182 86 9.002 23254 645190 266 -4378 2678 -13702 9842 -31234 974
Conclusions
We have shown two algorithms to convert a conjugate solution for cubic Fermat equation, in quadratic integers, into a non conjugate solution. We have used Alexander Aigner solutions form [3], to build these algorithms. Acknowledgements: we acknowledge Roman Orus for proposing the dissemination of this result.
References [1] 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. [2] 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 [3] 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) Email:
[email protected]
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