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.
"Nitpicking is the sincerest form of appreciation"
ReplyDeletethe part of line number 129 that reads
"(a2 + 3b2)3 = a6 + 9a4b2 + 9a2b4 + 27b6"
should be
"(a2 + 3b2)3 = a6 + 9a4b2 + 27a2b4 + 27b6"