Fermat’s Last Theorem (n = 3 and n = 4)

History

After the French mathematician Pierre de Fermat died in the seventeenth century, a note written by Fermat was discovered in which he conjectured that there were no nonzero integer solutions to the equation xⁿ + yⁿ = zⁿ for n > 2.  This proposition became known as Fermat’s Last Theorem. 

Fermat also wrote that he had “discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain,” but no proof could be found in any of his other notes.  At the time, though, this seemed trivial, because surely some other mathematician would find a proof for it.  But as years turned to decades, and decades turned to centuries, no general proof could be found.

The proof for Fermat’s Last Theorem became a Holy Grail for mathematics.  Like the Holy Grail, some mathematicians spent their lives trying to find a proof but came back empty-handed.  And like the Holy Grail, other mathematicians doubted its existence.  Perhaps Fermat’s Last Theorem was incorrect and so couldn’t be proved (although no counter-example could be found, either).  Perhaps Fermat didn’t actually have a proof, or perhaps he thought he had a proof but then realized later that it was incorrect and so never published it.  On the other hand, perhaps Fermat had a proof but there have been no mathematicians as great as him since to re-create his proof.  In any event, by the beginning of the twentieth century finding a general proof for Fermat’s Last Theorem became sweeter than just mathematical notoriety, because an industrialist named Wolfskehl bequeathed a prize of 100,000 German Marks to the first mathematician who could prove it.

Fermat’s Last Theorem: The Holy Grail of Mathematics

In the meantime, specific proofs were found for Fermat’s Last Theorem.  A proof from Fermat himself was discovered for the specific case of n = 4 (that there were no nonzero integer solutions to the equation x⁴ + y⁴ = z⁴).  Euler proved Fermat’s Last Theorem for n = 3 in 1770, over a hundred years after Fermat’s death.  Dirichlet and Legendre proved Fermat’s Last Theorem for n = 5 in 1825, nearly two hundred years after Fermat’s death. 

More and more specific proofs were found for greater and greater exponents, but it was not until 1995, over three hundred years after Fermat’s death, that Wiles published a general proof that included all exponents greater than two.  His proof is over one hundred pages long, and uses modern methods and techniques not available to Fermat (and beyond the scope of this article), so most mathematicians agree that this is probably not the proof Fermat had in mind back in the seventeenth century (if he even had one).

Wiles

Difficulties

One of the reasons why Fermat’s Last Theorem is so difficult to prove is because it has to be true for a certain set of numbers but false for another set of numbers.  First of all, Fermat’s Last Theorem only applies for nonzero integer solutions.  However, integer solutions to xⁿ + yⁿ = zⁿ do exist for all exponents n as long as one of x, y, or z is equal to zero, for example, 1³ + 0³ = 1³ or 2⁴ + 0⁴ = 2⁴.

Secondly, Fermat’s Last Theorem also only applies for exponents greater than two.  Nonzero integer solutions to xⁿ + yⁿ = zⁿ do exist for n = 1 and n = 2.  When n = 1, the formula becomes x + y = z, which has infinitely many nonzero integer solutions, for example, 3 + 4 = 7 or 5 + 12 = 17.  When n = 2, the formula becomes x² + y² = z², which is the Pythagorean Theorem, and since it can be proved that all solutions to the Pythagorean Theorem are x = p² – q², y = 2pq, and y = p² + q² for any p and q (see here for a proof), an infinitely many combinations of nonzero integers p and q can be picked to generate nonzero integer solutions.  For example, if p = 2 and q = 1, then x = p² – q² = 2² – 1² = 3, y = 2pq = 2(2)(1) = 4, and z = p² + q² = 2² + 1² = 5, so x² + y² = z² is 3² + 4² = 5².  Or for another example, if p = 3 and q = 2, then x = p² – q² = 3² – 2² = 5, y = 2pq = 2(3)(2) = 12, and z = p² + q² = 3² + 2² = 13, so x² + y² = z² is 5² + 12² = 13².

Fermat’s Last Theorem for n = 3

Recall from above that Euler proved Fermat’s Last Theorem for n = 3 (that there are no nonzero integer solutions to x³ + y³ = z³) in 1770.  A few different ways to prove Fermat’s Last Theorem for n = 3 have been found, and a full proof, which is loosely based on the proof found on Freeman’s blog, can be found here. 

Summarizing the proof briefly, first assume that there are positive integer solutions to x³ + y³ = z³.  Then there must exist a cube in the form of 2p(p² + 3q²) where p and q are relatively prime and where the greatest common factor of 2p and p² + 3q² is 1 or 3.  In either case, you can use the fact that all odd factors of p² + 3q² must also have that same form (proof for that here) to prove that there exists smaller positive integers x’, y’, and z’ in which x’³ + y’³ = z’³.  Since we can apply the same argument infinitely to obtain smaller and smaller positive integers, there is a contradiction by the method of infinite descent, and so we reject our assumption that there are positive solutions to x³ + y³ = z³.  We can also reject that there are any negative solutions to x³ + y³ = z³ because the formula can be rearranged to an all positive integer solution still in the same form.  Since there are no positive or negative solutions to x³ + y³ = z³, there are no nonzero integer solutions to x³ + y³ = z³.

Fermat’s Last Theorem for n = 4

Also recall from above that a proof for n = 4 (that there are no nonzero integer solutions to x⁴ + y⁴ = z⁴) was given by Fermat himself.  The specific case for n = 4 is actually easier than proving the specific case for n = 3, because a fourth power is a square of a square, so a Pythagorean solution can be found twice in such a way to produce a contradiction once again by the method of infinite descent.  A full proof for Fermat’s Last Theorem for n = 4 can be found here.

Fermat’s Last Theorem for n ≥ 5

Proving Fermat’s Last Theorem for n = 3 proves Fermat’s Last Theorem for n for all multiples of 3, and likewise proving Fermat’s Last Theorem for n = 4 proves Fermat’s Last Theorem for n for all multiples of 4.  For example, to prove that there are no nonzero integer solutions for n for all multiples of 3, or n = 3m, one can simply express x^3m + y^3m = z^3m as (x^m)³ + (y^m)³ = (z^m)³, a special case for n = 3; and to prove that there are no nonzero integer solutions for n for all multiples of 4, or n = 4m, one can simply express x^4m + y^4m = z^4m as (x^m)⁴ + (y^m)⁴ = (z^m)⁴, a special case of n = 4.  So a general proof for Fermat’s Last Theorem only requires a proof for n of all prime numbers greater than two.

It is tempting to use the proofs for n = 3 or n = 4 and apply them to the proofs for higher exponents, but unfortunately both proofs for n = 3 and n = 4 exploit properties that are unique for its exponent.  The proof for n = 3 uses a unique property that all odd factors of p² + 3q² must also have that same form.  But using the same logic as the proof for n = 3 on n = 5, we would need all odd factors of p⁴ + 10p²q² + 5q⁴ to have the same form, but unfortunately this is simply not true, and the same can be said for higher exponents.  Furthermore, the proof for n = 4 uses the unique property that a fourth power is a square of a square, which is not true of any primes greater than two.  So a specific proof for any prime n ≥ 5 must use a completely different (and more difficult) approach altogether.

Conclusion

Even though Fermat’s Last Theorem was proved by Wiles in 1995, some mysteries remain.  Did Fermat really have a legitimate general proof for his theorem, or was he mistaken?  Will a general proof be found that is shorter or uses a traditional approach?  Perhaps it will take another three centuries to answer these questions.  If only Fermat’s margin was bigger!

Fermat’s Last Theorem (Proof for n = 4)

There are no nonzero integer solutions to x⁴ + y⁴ = z⁴

001	Assume x4 + y4 = z4, where x, y and z are nonzero integers.
002	Letting w = z2, we have x4 + y4 = w2.
003	Dividing out any common factors, there exists an X, Y, and W such that X4 + Y4 = W2 in which X, Y, and W are relatively prime.
004	Then (X2)2 + (Y2)2 = W2
005	Then there exists a Pythagorean solution for (X2)2 + (Y2)2 = W2, which is X2 = p2 – q2, Y2 = 2pq, and W = p2 + q2, where p, q and X are relatively prime.
006	Then there also exists a second Pythagorean solution for X2 = p2 – q2, which is X = m2 – n2, p = m2 + n2, and q = 2mn, where m and n are relatively prime.
007	Assume p and m are not relatively prime:
008	Then p = fP and m = fM
009	And n is also divisible by f since n2 = p – m2 = fP – (fM)2 = fP – f2M2 = f(P – fM2)
010	Then n and m have a common factor f.
011	But this contradicts that m and n are relatively prime.
012	So p and m are relatively prime.
013	Assume p and n are not relatively prime:
014	Then p = fP and n = fN
015	And m is also divisible by f since m2 = p + n2 = fP + (fN)2 = fP + f2N2 = f(P + fN2)
016	Then n and m have a common factor f.
017	But this contradicts that m and n are relatively prime.
018	So p and n are relatively prime.
019	Therefore, since m and n are relatively prime, and p and m are relatively prime, and p and n are relatively prime, that means p, m, and n are relatively prime.
020	Now  Y2 = 4pmn, because Y2 = 2pq = 2p(2mn) = 4pmn.
021	Since Y2 = 4pmn and p, m, and n are relatively prime, p, m, and n are squares, so p = w’2, m = x’2, and n = y’2
022	Substituting into p = m2 + n2, we have w’2 = (x’2)2 + (y’2)2, or x’4 + y’4 = w’2.
023	Now x’ < w since x’ is a factor of m which is a factor of q which (because W = p2 + q2) is less than W which a factor of w
024	And y’ < w since y’ is a factor of n which is a factor of q which (because W = p2 + q2) is also less than W which is a factor of w
025	And w’ < w since w’ is a factor of p which (because W = p2 + q2) is also less than W which is a factor of w
026	So there exists integers x’, y’, and w’ such that x’4 + y’4 = w’2 and 1 ≤ x’ < w, 1 ≤ y’ < w, and 1 ≤ w’ < w.
027	Therefore, if x4 + y4 = w2, there must exist other positive integers x’, y’, w’ such that 1 ≤ x’ < w, 1 ≤ y’ < w, and 1 < w’ < w in which x’4 + y’4 = w’2.
028	But we can use the same argument to prove the existence of more positive integers x”, y”, w” such that 1 ≤ x” < w’, 1 ≤ y” < w’, and 1 < w” < w’ in which x”4 + y”4 = w”2, and another x’”, y’”, w’”, and so on, so we have a contradiction by the method of infinite descent.
029	Therefore, there are no nonzero integer solutions to x4 + y4 = z4.

Q.E.D.

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.