Monday, July 13, 2015

Fermat’s Last Theorem (Proof for n = 3)

There are no nonzero integer solutions to x3 + y3 = z3

001
Assume x3 + y3 = z3, where x, y and z are positive integers.
002
Then x < z and y < z.
003
Dividing out any common factors, there exists an X, Y, and Z such that X3 + Y3 = Z3 in which X, Y, and Z are relatively prime and X < Z and Y < Z.
004
Let f be the greatest common factor of x and y.
005
Then x = fX and y = fY.
006
And f divides into z since z3 = x3 + y3 = (fX)3 + (fY)3 = f3X3 + f3Y3 = f3(X3 + Y3)
007
So let z = fZ.
008
Therefore x3 + y3 = z3 is (fX)3 + (fY)3 = (fZ)3 which is f3X3 + f3Y3 = f3Z3 which is X3 + Y3 = Z3, where X and Y are relatively prime.
009
Assume X and Z are not relatively prime:
010
Then X = gX’ and Z = gZ’ where g > 1
011
Then g divides into Y since Y3 = Z3 – X3 = (gZ’)3 – (gX’)3 = g3Z’3 – g3X’3 = g3(Z’3 – X’3)
012
But this contradicts that X and Y are relatively prime.
013
So X and Z are relatively prime.
014
Assume Y and Z are not relatively prime:
015
Then Y = gY’ and Z = gZ’ where g > 1
016
Then g divides into X since X3 = Z3 – Y3 = (gZ’)3 – (gY’)3 = g3Z’3 – g3Y’3 = g3(Z’3 – Y’3)
017
But this contradicts that X and Y are relatively prime.
018
So Y and Z are relatively prime.
019
Since x < z then fX < fZ and X < Z.
020
Since y < z then fY < fZ and Y < Z.
021
Exactly one of X, Y, and Z are even, and the other two are odd.
022
Assume more than one of X, Y, and Z are even:
023
Then two of X, Y, and Z are divisible by 2.
024
But this contradicts that X, Y, and Z are relatively prime.
025
So it cannot be the case that more than one of X, Y, and Z are even.
026
Assume none of X, Y, and Z are even:
027
Then X, Y, and Z are all odd.
028
Since an odd number cubed is always an odd number, then X3, Y3, and Z3 are all odd.
029
But since an odd number plus an odd number is always an even number, Z3 which is X3 + Y3 must be an even number.
030
But Z3 cannot be an odd number and an even number at the same time, so there is a contradiction.
031
So it cannot be the case that none of X, Y, and Z are even.
032
Since more than one of X, Y, and Z cannot be even, and less than one of X, Y, and Z cannot be even, then exactly one of X, Y, and Z are even.
033
There exists integers u, p, and q such that u3 = 2p(p2 + 3q2) in which p is a positive integer and in which p and q are relatively prime and have opposite parity.
034
Since exactly one of X, Y, and Z are even, either X is even and Y and Z are odd, or Y is even and X and Z are odd, or Z is even and X and Y are odd.
035
Case 1: X is even and Y and Z are odd:
036
Then Z – Y is an odd number minus an odd number, which is even, so let Z – Y = 2p, where p is a positive integer.
037
And Z + Y is odd number plus an odd number, which is also even, so let Z + Y = 2q, where q is a positive integer.
038
Then Z = q + p, because Z = ½(2Z) = ½(Z + Z) = ½(Z + Z + Y – Y) = ½(Z + Y + Z – Y) = ½((Z + Y) + (Z – Y)) = ½(2q + 2p) = q + p.
039
And Y = q – p, because Y = ½(2Y) = ½(Y + Y) = ½(Y + Y + Z – Z) = ½(Z + Y – Z + Y) = ½((Z + Y) – (Z – Y)) = ½(2q – 2p) = q – p.
040
And if u = X, u3 = 2p(p2 + 3q2), because u3 = X3 = Z3 – Y3 = (q + p)3 – (q – p)3 = (q3 + 3q2p + 3qp2 + p3) – (q3 – 3q2p + 3qp2 – p3) = 6q2p + 2p3 = 2p(3q2 + p2) = 2p(p2 + 3q2).
041
Assume p and q are not relatively prime:
042
Then p = fP and q = fQ where f > 1.
043
And Z is divisible by f because Z = q + p = fQ + fP = f(Q + P).
044
And Y is divisible by f because Y = q – p = fQ – fP = f(Q – P).
045
So Z and Y have a common factor f.
046
But this contradicts that Z and Y are relatively prime.
047
So p and q are relatively prime.
048
Assume p and q are both even:
049
Then both p and q are divisible by 2.
050
But this contradicts that p and q are relatively prime.
051
So p and q cannot be both even.
052
Assume p and q are both odd:
053
Then Z = q + p is even, because an odd plus an odd makes an even.
054
But this contradicts that Z is an odd.
055
So p and q cannot be both odd.
056
Since p and q cannot be both even nor can p and q both be odd, p and q must have opposite parity.
057
So there exists integers u, p, and q such that u3 = 2p(p2 + 3q2) in which p is a positive integer and in which p and q are relatively prime and have opposite parity.
058
Case 2: Y is even and X and Z are odd:
059
Then Z – X is an odd number minus an odd number, which is even, so let Z – X = 2p, where p is a positive integer
060
And Z + X is odd number plus an odd number, which is also even, so let Z + X = 2q, where q is a positive integer.
061
Then Z = q + p, because Z = ½(2Z) = ½(Z + Z) = ½(Z + Z + X – X) = ½(Z + X + Z – X) = ½((Z + X) + (Z – X)) = ½(2q + 2p) = q + p.
062
And X = q – p, because X = ½(2X) = ½(X + X) = ½(X + X + Z – Z) = ½(Z + X – Z + X) = ½((Z + X) – (Z – X)) = ½(2q – 2p) = q – p.
063
And if u = Y, u3 = 2p(p2 + 3q2), because u3 = Y3 = Z3 – X3 = (q + p)3 – (q – p)3 = (q3 + 3q2p + 3qp2 + p3) – (q3 – 3q2p + 3qp2 – p3) = 6q2p + 2p3 = 2p(3q2 + p2) = 2p(p2 + 3q2).
064
Assume p and q are not relatively prime:
065
Then p = fP and q = fQ where f > 1.
066
And Z is divisible by f because Z = q + p = fQ + fP = f(Q + P).
067
And X is divisible by f because X = q – p = fQ – fP = f(Q – P).
068
So Z and X have a common factor f.
069
But this contradicts that Z and X are relatively prime.
070
So p and q are relatively prime.
071
Assume p and q are both even:
072
Then both p and q are divisible by 2.
073
But this contradicts that p and q are relatively prime.
074
So p and q cannot be both even.
075
Assume p and q are both odd:
076
Then Z = q + p is even, because an odd plus an odd makes an even.
077
But this contradicts that Z is an odd.
078
So p and q cannot be both odd.
079
Since p and q cannot be both even nor can p and q both be odd, p and q must have opposite parity.
080
So there exists integers u, p, and q such that u3 = 2p(p2 + 3q2) in which p is a positive integer and in which p and q are relatively prime and have opposite parity.
081
Case 3: Z is even and X and Y are odd:
082
And Y + X is odd number plus an odd number, which is also even, so let Y + X = 2p, where p is a positive integer.
083
Then Y – X is an odd number minus an odd number, which is even, so let Y – X = 2q, where q is an integer.
084
Then Y = p + q, because Y = ½(2Y) = ½(Y + Y) = ½(Y + Y + X – X) = ½(Y + X + Y – X) = ½((Y + X) + (Y – X)) = ½(2p + 2q) = p + q.
085
And X = p – q, because X = ½(2X) = ½(X + X) = ½(X + X + Y – Y) = ½(Y + X – Y + X) = ½((Y + X) – (Y – X)) = ½(2p – 2q) = p – q.
086
And if u = Z, u3 = 2p(p2 + 3q2), because u3 = Z3 = X3 + Y3 = (p – q)3 + (p + q)3 = (p3 – 3p2q + 3pq2 – q3) + (p3 + 3p2q + 3pq2 + q3) = 2p3 + 6pq2 = 2p(p2 + 3q2).
087
Assume p and q are not relatively prime:
088
Then p = fP and q = fQ where f > 1.
089
And X is divisible by f because X = p + q = fP + fQ = f(P + Q).
090
And Y is divisible by f because Y = p – q = fP – fQ = f(P – Q).
091
So X and Y have a common factor f.
092
But this contradicts that X and Y are relatively prime.
093
So p and q are relatively prime.
094
Assume p and q are both even:
095
Then both p and q are divisible by 2.
096
But this contradicts that p and q are relatively prime.
097
So p and q cannot be both even.
098
Assume p and q are both odd:
099
Then Y = q + p is even, because an odd plus an odd makes an even.
100
But this contradicts that Y is an odd.
101
So p and q cannot be both odd.
102
Since p and q cannot both be even, nor can p and q both be odd, p and q must have opposite parity.
103
So there exists integers u, p, and q such that u3 = 2p(p2 + 3q2) in which p is a positive integer and in which p and q are relatively prime and have opposite parity.
104
In all three cases, there exists integers u, p, and q such that u3 = 2p(p2 + 3q2) in which p is a positive integer and in which p and q are relatively prime and have opposite parity.
105
Since p and q have opposite parity, p2 + 3q2 is odd
106
Case 1: p is even and q is odd:
107
Then p2 is even because an even times an even is an even.
108
And 3q2 is odd because an odd times an odd times an odd is an odd.
109
So p2 + 3q2 is odd because an even plus an odd is an odd.
110
Case 2: p is odd and q is even:
111
Then p2 is odd because an odd times an odd is an odd.
112
And 3q2 is even because an odd times an even times an even is an even.
113
And p2 + 3q2 is odd because an odd plus and even is odd.
114
In either case, p2 + 3q2 is odd.
115
The greatest common factor of 2p and p2 + 3q2 is either 1 or 3.
116
Assume p and p2 + 3q2 have a common factor f > 3:
117
Then p = fP and p2 + 3q2 = fN
118
Then q is divisible by f as well because 3q2 = fN – p2 = fN – (fP)2 = fN – f2P2 = f(N – fP2)
119
So p and q have a common factor f > 3
120
But this contradicts that p and q are relatively prime.
121
So p and p2 + 3q2 do not have a common factor f > 3.
122
Since p2 + 3q2 is odd, it cannot have a factor of 2.
123
Since p and p2 + 3q2 do not have a common factor f > 3, and p2 + 3q2 does not have a factor of 2, then 2p and p2 + 3q2 do not have a common factor f > 3 nor a common factor of 2.
124
In other words, the greatest common factor of 2p and p2 + 3q2 is either 1 or 3.
125
Case 1: The greatest common factor of 2p and p2 + 3q2 is 1:
126
Since u3 = 2p(p2 + 3q2) and the greatest common factor of 2p and p2 + 3q2 is 1, then both 2p and p2 + 3q2 must be cubes.
127
Since p and q are relatively prime, then every odd factor of p2 + 3q2 must also have the same form, because of the theorem for a2 + 3b2 (see here)
128
Since p2 + 3q2 is a cube and an odd number, its cube root must also be odd, and as an odd factor of p2 + 3q2 it must also have the same form.  So there must exist positive integers a and b such that p2 + 3q2 = (a2 + 3b2)3.
129
So p = a3 – 9ab2 and q = 3a2b – 3b3, because p2 + 3q2 = (a2 + 3b2)3 = a6 + 9a4b2 + 9a2b4 + 27b6 = a6 + 27a4b2 – 18a4b2 + 81a2b4 – 54a2b4 + 27b6 = a6 – 18a4b2 + 81a2b4 + 27a4b2 – 54a2b4 + 27b6 = a6 – 18a4b2 + 81a2b4 + 3(9a4b2 – 18a2b4 + 9b6) = (a3 – 9ab2)2 + 3(3a2b – 3b)2.
130
So 2p = 2a(a – 3b)(a + 3b) because 2p = 2(a3 – 9ab2) = 2a(a – 3b)(a + 3b).
131
Now 2a, a – 3b and a + 3b are relatively prime.
132
Assume a and b are not relatively prime:
133
Then a = fA and b = fB.
134
So p is divisible by f because p = a3 – 9ab2 = (fA)3 – 9(fA)(fB)2 = f3A3 – 9f3AB2 = f3(A3 – 9AB2)
135
And q is divisible by f because q = 3a2b – 3b3 = 3(fA)2(fB) – 3(fB)3 = 3f3A2B – 3f3B3 = f3(3A2B – 3B3)
136
So p and q have a common factor f.
137
But this contradicts that p and q are relatively prime.
138
So a and b are relatively prime.
139
Assume a and b are both even:
140
Then a and b are both divisible by 2.
141
But this contradicts that a and b are relatively prime.
142
So a and b cannot both be even.
143
Assume a and b are both odd:
144
Then p is even because p = a3 – 9ab2 which is an odd minus an odd which is even.
145
And q is even because q = 3a2b – 3b3 which is an odd minus an odd which is even.
146
Then p and q are both divisible by 2
147
But this contradicts that p and q are relatively prime
148
So a and b cannot both be odd.
149
Since a and b cannot both be even, nor can a and b both be odd, a and b must have opposite parity.
150
Since a and b have opposite parity, a – 3b and a + 3b are both odd
151
Case 1: a is even, b is odd:
152
Then a – 3b is an even minus an odd which is an odd.
153
And a + 3b is an even plus an odd which is also an odd.
154
Case 2: a is odd, b is even:
155
Then a – 3b is an odd minus an even which is an odd.
156
And a + 3b is an odd plus an even which is also an odd.
157
Either way, a – 3b and a + 3b are both odd.
158
Assume a is divisible by 3:
159
Then a = 3A.
160
And p is divisible by 3 because p = a3 – 9ab2 = a(a2 – 9b2) = 3A(a2 – 9b2).
161
And q is divisible by 3 because q = 3a2b – 3b3 = 3(a2b – b3).
162
So both p and q have a common factor of 3.
163
But this contradicts that p and q are relatively prime.
164
So a is not divisible by 3.
165
Assume a and a – 3b have a common factor f > 3:
166
Then a = fA and a – 3b = fN
167
Then b is divisible by f as well because 3b = a – fN = fA – fN = f(A – N).
168
So a and b have a common factor f > 3
169
But this contradicts that a and b are relatively prime.
170
So a and a – 3b do not have a common factor f > 3.
171
Since a – 3b is odd, 2a and a – 3b does not have a common factor f > 3, either.
172
Since a is not divisible by 3, neither is 2a, and so 2a and a – 3b do not have a common factor of 3.
173
Since a – 3b is odd, 2a and a – 3b do not have a common factor of 2.
174
Since 2a and a – 3b do not have a common factor f > 3, or 3, or 2, this means that 2a and a – 3b must be relatively prime.
175
Assume 2a and a + 3b are not relatively prime:
176
Then 2a = fM and a + 3b = fN
177
Then a – 3b is also divisible by f, because a – 3b = 2a – a – 3b = 2a – (a + 3b) = fM – fN = f(M – N)
178
But this contradicts that 2a and a – 3b are relatively prime.
179
So 2a and a + 3b must also be relatively prime.
180
Assume a – 3b and a + 3b are not relatively prime:
181
Then a – 3b = fM and a + 3b = fN.
182
Then 2a is also divisible by f, because 2a = a + a = a + a + 3b – 3b = a – 3b + a + 3b = fM + fN = f(M + N)
183
But this contradicts that 2a and a – 3b are relatively prime.
184
So a – 3b and a + 3b must also be relatively prime.
185
Since 2a and a – 3b are relatively prime, and 2a and a + 3b are relatively prime, and a – 3b and a + 3b are relatively prime, then 2a, a – 3b, and a + 3b are relatively prime.
186
Since 2p is a cube and 2p = 2a(a – 3b)(a + 3b) and 2a, a – 3b, and a + 3b are relatively prime, then 2a, a – 3b and a + 3b are all cubes as well, so x’3 = a – 3b, y’3 = a + 3b and z’3 = 2a.
187
Then x’3 + y’3 = z’3, because x’3 + y’3 = (a – 3b) + (a + 3b) = 2a = z’3.
188
Since x’3y’3z’3 = 2p and 2p is a factor of either X3, Y3, or Z3 which is a factor of either x3, y3, or z3, x’, y’ and z’ are all smaller than z.
189
Since a and b are positive integers and y’ = a + 3b and z’ = 2a, y’ and z’ are positive integer greater than 1.
190
Since p, y’ and z’ are positive and x’3y’3z’3 = 2p, x’ is also a positive integer greater than 1.
191
So there exists integers x’, y’, and z’ such that x’3 + y’3 = z’3 and 1 ≤ x’ < z, 1 ≤ y’ < z, and 1 ≤ z’ < z.
192
Case 2: The greatest common factor of 2p and p2 + 3q2 is 3:
193
Then 2p is divisible by 3, and p is divisible by 3, so let p = 3r.
194
Since p and q are relatively prime and p = 3r, q and r are also relatively prime.
195
Since p and q have opposite parity and p = 3r, q and r also have opposite parity.
196
Also, u3 = 18r(q2 + 3r2) because u3 = 2p(p2 + 3q2) = 2(3r)((3r)2 + 3q2) = 6r(9r2 + 3q2) = 18r(3r2 + q2) = 18r(q2 + 3r2)
197
Since q and r have opposite parity, q2 + 3r2 is odd
198
Case 1: q is even and r is odd:
199
Then q2 is even because an even times an even is an even.
200
And 3r2 is odd because an odd times an odd times an odd is an odd.
201
So q2 + 3r2 is odd because an even plus an odd is an odd.
202
Case 2: q is odd and r is even:
203
Then q2 is odd because an odd times an odd is an odd.
204
And 3r2 is even because an odd times an even times an even is an even.
205
And q2 + 3r2 is odd because an odd plus and even is odd.
206
In either case, q2 + 3r2 is odd.
207
Also, 18r and q2 + 3r2 are relatively prime.
208
Since p = 3r and p and q are relatively prime, q is not divisible by 3.
209
Assume q2 + 3r2 is divisible by 3.
210
Then q2 + 3r2 = 3s
211
And q is divisible by 3 since q2 = 3s – 3r2 = 3(s – r2)
212
But this contradicts that q is not divisible by 3.
213
So q2 + 3r2 is not divisible by 3.
214
Assume r and q2 + 3r2 have a common factor f > 3:
215
Then r = fR and q2 + 3r2 = fN
216
Then q is divisible by f because q2 = fN – 3r2 = fN – 3(fR)2 = fN – 3f2R2 = f(N – 3fR2)
217
Then q and r have a common factor f.
218
But this contradicts that q and r are relatively prime.
219
So r and q2 + 3r2 do not have a common factor f > 3.
220
Since r and q2 + 3r2 do not have a common factor f > 3 and q2 + 3r2 is not divisible by 3, this means that 3r and q2 + 3r2 do not have a common factor f > 3.
221
Since q2 + 3r2 is not divisible by 3, 3r and q2 + 3r2 do not have a common factor of 3.
222
Since q2 + 3r2 is odd, 3r and q2 + 3r2 do not have a common factor of 2.
223
Since 3r and q2 + 3r2 do not have a common factor f > 3, or 3, or 2, this means that 3r and q2 + 3r2 must be relatively prime.
224
Since u3 = 18r(q2 + 3r2) and the greatest common factor of 18r and q2 + 3r2 is 1, then both 18r and q2 + 3r2 must be cubes.
225
Since q and r are relatively prime, then every odd factor of q2 + 3r2 must also have the same form, because of the theorem for a2 + 3b2 (see here)
226
Since q2 + 3r2 is a cube and an odd number, its cube root must also be odd, and as an odd factor of q2 + 3r2 it must also have the same form.  So there must exist positive integers a and b such that q2 + 3r2 = (a2 + 3b2)3.
227
So q = a3 – 9ab2 and r = 3a2b – 3b3, because q2 + 3r2 = (a2 + 3b2)3 = a6 + 9a4b2 + 9a2b4 + 27b6 = a6 + 27a4b2 – 18a4b2 + 81a2b4 – 54a2b4 + 27b6 = a6 – 18a4b2 + 81a2b4 + 27a4b2 – 54a2b4 + 27b6 = a6 – 18a4b2 + 81a2b4 + 3(9a4b2 – 18a2b4 + 9b6) = (a3 – 9ab2)2 + 3(3a2b – 3b)2.
228
So 18r = 33(2b)(a – b)(a + b) because 18r = 18(3a2b – 3b3) = 18(3b)(a2 – b2) = 54b(a – b)(a + b) = 33(2b)(a – b)(a + b).
229
Now 2b, a – b and a + b are relatively prime.
230
Assume a and b are not relatively prime:
231
Then a = fA and b = fB.
232
So q is divisible by f because q = a3 – 9ab2 = (fA)3 – 9(fA)(fB)2 = f3A3 – 9f3AB2 = f3(A3 – 9AB2)
233
And r is divisible by f because r = 3a2b – 3b3 = 3(fA)2(fB) – 3(fB)3 = 3f3A2B – 3f3B3 = f3(3A2B – 3B3)
234
So q and r have a common factor f.
235
But this contradicts that q and r are relatively prime.
236
So a and b are relatively prime.
237
Assume a and b are both even:
238
Then a and b are both divisible by 2.
239
But this contradicts that a and b are relatively prime.
240
So a and b cannot both be even.
241
Assume a and b are both odd:
242
Then q is even because p = a3 – 9ab2 which is an odd minus an odd which is even.
243
And r is even because q = 3a2b – 3b3 which is an odd minus an odd which is even.
244
Then q and r are both divisible by 2
245
But this contradicts that q and r are relatively prime
246
So a and b cannot both be odd.
247
Since a and b cannot both be even, nor can a and b both be odd, a and b must have opposite parity.
248
Since a and b have opposite parity, a – b and a + b are both odd
249
Case 1: a is even, b is odd:
250
Then a – b is an even minus an odd which is an odd.
251
And a + b is an even plus an odd which is also an odd.
252
Case 2: a is odd, b is even:
253
Then a – b is an odd minus an even which is an odd.
254
And a + b is an odd plus an even which is also an odd.
255
Either way, a – b and a + b are both odd.
256
Assume b and a – b are not relatively prime:
257
Then b = fB and a – b = fN
258
Then a is divisible by f as well because a = b + fN = fB + fN = f(B + N).
259
So a and b have a common factor f
260
But this contradicts that a and b are relatively prime.
261
So b and a – b are relatively prime.
262
Since a – b is odd and b and a – b are relatively prime, 2b and a – b are relatively prime, too.
263
Assume 2b and a + b are not relatively prime:
264
Then 2b = fM and a + b = fN
265
Then a – b is also divisible by f, because a – b = a + b – 2b = (a + b) – 2b = fN – fM = f(N – M)
266
But this contradicts that 2b and a – b are relatively prime.
267
So 2b and a + b must also be relatively prime.
268
Assume a – b and a + b are not relatively prime:
269
Then a – b = fM and a + b = fN.
270
Then 2b is also divisible by f, because 2b = b + b = b + b + a – a = a + b – a + b = (a + b) – (a – b) = fN – fM = f(N – M).
271
But this contradicts that 2b and a – b are relatively prime.
272
So a – b and a + b must also be relatively prime.
273
Since 2b and a – b are relatively prime, and 2b and a + b are relatively prime, and a – b and a + b are relatively prime, then 2b, a – b, and a + b are relatively prime.
274
Since 18r is a cube and 18r = 33(2b)(a – b)(a + b) and 2b, a – b, and a + b are relatively prime, then 2b, a – b and a + b are all cubes as well, so x’3 = a – b, y’3 = 2b, and z’3 = a + b.
275
Then x’3 + y’3 = z’3, because x’3 + y’3 = (a – b) + (2b) = a + b = z’3.
276
Since 33x’3y’3z’3 = 18r and 18r is a factor of either X3, Y3, or Z3 which is a factor of either x3, y3, or z3, x’, y’ and z’ are all smaller than z.
277
Since a and b are positive integers and y’ = 2b and z’ = a + b, y’ and z’ are positive integer greater than 1.
278
Since p, y’ and z’ are positive and 33x’3y’3z’3 = 18r = 6p, x’ is also a positive integer greater than 1.
279
So there exists integers x’, y’, and z’ such that x’3 + y’3 = z’3 and 1 ≤ x’ < z, 1 ≤ y’ < z, and 1 ≤ z’ < z.
280
In either case, there exists integers x’, y’, and z’ such that x’3 + y’3 = z’3 and 1 ≤ x’ < x, 1 ≤ y’ < y, and 1 ≤ z’ < z.
281
Therefore, if x3 + y3 = z3, there must exist other positive integers x’, y’, z’ such that 1 ≤ x’ < z, 1 ≤ y’ < z, and 1 < z’ < z in which x’3 + y’3 = z’3.
282
But we can use the same argument to prove the existence of more positive integers x”, y”, z” such that 1 ≤ x” < z’, 1 ≤ y” < z’, and 1 < z” < z’ in which x”3 + y”3 = z”3, and another x’”, y’”, z’”, and so on, so we have a contradiction by the method of infinite descent.
283
So there are no positive integer solutions to x3 + y3 = z3
284
Any negative solution to x3 + y3 = z3 can be rearranged to an all positive solution also in the same form.
285
Case 1: there is exactly one negative solution to x3 + y3 = z3
286
Case a: x is negative, y and z are positive:
287
Then let X = -x, Y = z, and Z = y, where X, Y, and Z are positive integers
288
So x3 + y3 = z3 becomes (-X)3 + Z3 = Y3 or X3 + Y3 = Z3
289
Case b: y is negative, x and z are positive:
290
Then let X = z, Y = -y, and Z = x, where X, Y, and Z are positive integers
291
So x3 + y3 = z3 becomes Z3 + (-Y)3 = X3 or X3 + Y3 = Z3
292
Case c: z is negative, x and y are positive:
293
This is impossible, since x3 + y3 = z3 and a positive plus a positive must be a positive, not a negative.
294
Case 2: there are exactly two negative solutions to x3 + y3 = z3
295
Case a: x and y are negative, z is positive:
296
This is impossible, since x3 + y3 = z3 and a negative plus a negative must be a negative, not a positive.
297
Case b: x and z are negative, y is positive:
298
Then let Z = -x, Y = y, and X = -z, where X, Y, and Z are positive integers
299
So x3 + y3 = z3 becomes (-Z)3 + Y3 = (-X)3 or X3 + Y3 = Z3
300
Case c: y and z are negative, x is positive:
301
Then let X = x, Y = -z, and Z = -y, where X, Y, and Z are positive integers
302
So x3 + y3 = z3 becomes X3 + (-Z)3 = (-Y)3 or X3 + Y3 = Z3
303
Case 3: there are exactly three negative solutions to x3 + y3 = z3
304
Then let X = -x, Y = -y, and Z = -z, where X, Y, and Z are positive integers
305
So x3 + y3 = z3 becomes (-X)3 + (-Y)3 = (-Z)3 or X3 + Y3 = Z3
306
In all cases, there exist positive integers X, Y, and Z such that X3 + Y3 = Z3.
307
Since any negative solution to x3 + y3 = z3 can be rearranged to an all positive solution also in the same form, and there are no positive integer solutions to x3 + y3 = z3, then there are no negative integer solutions for x3 + y3 = z3, either.
308
Therefore, there are no nonzero integer solutions to x3 + y3 = z3

Q.E.D.

1 comment:

  1. "Nitpicking is the sincerest form of appreciation"

    the part of line number 129 that reads
    "(a2 + 3b2)3 = a6 + 9a4b2 + 9a2b4 + 27b6"
    should be
    "(a2 + 3b2)3 = a6 + 9a4b2 + 27a2b4 + 27b6"

    ReplyDelete