Fermat’s Last Theorem (Proof of Theorem for a^2 + 3b^2)

If a and b are relatively prime, then every odd factor of a² + 3b² must also have the same form.

001	Let x be a positive odd factor and let f be the other factor such that a2 + 3b2 = xf.
002	Assume that x is not in the form a2 + 3b2:
003	x ≠ 1 because 1 is in the form of a2 + 3b2 (1 = 12 + 3(0)2)
004	Let a = mx ± c and b = nx ± d, where 0 < c < ½x and 0 < d < ½x
005	Then a2 + 3b2 = (mx ± c)2 + 3(nx ± d)2 = m2x2 ± 2cmx + c2 + 3(n2x2 ± 2dnx + d2) = x(m2x ± 2cm + 3n2x ± 6dn) + c2 + 3d2 = xF + c2 + 3d2.
006	Then x divides into c2 + 3d2, since c2 + 3d2 = a2 + 3b2 – xF = xf – xF = x(f – F).
007	Let y = f – F.  Then xy = c2 + 3d2.
008	Since c < ½x and d < ½x, xy < (½x)2 + 3(½x)2 or xy < x2, so y < x.
009	Assume that all the odd factors of y have the form a2 + 3b2:
010	If y is even, then it is divisible by 4:
011	Assume y is even.
012	Since y is even and xy = c2 + 3d2, c2 + 3d2 is even.
013	In order for c2 + 3d2 to be even, c and d must have the same parity (either both or even or both or odd).
014	Case 1: both c and d are even:
015	Let c = 2C and d = 2D.
016	Then 4 divides into c2 + 3d2, since c2 + 3d2 = (2C)2 + 3(2D)2  = 4C2 + 12D2 = 4(C2 + 3D2).
017	Case 2: both c and d are odd:
018	Let c = 2C + 1 and d = 2D + 1.
019	Then 4 divides into c2 + 3d2, since c2 + 3d2 = (2C + 1)2 + 3(2D + 1)2  = 4C2 + 4C + 1 + 3(4D2 + 4D + 1) = 4C2 + 4C + 1 + 12D2 + 12D + 3 = 4C2 + 4C + 12D2 + 12D + 4 = 4(C2 + C + 3D2 + 3D + 1).
020	Either way, 4 divides into c2 + 3d2.
021	Since 4 divides into c2 + 3d2 and x is odd and xy = c2 + 3d2, 4 must divide into y as well.
022	If c2 + 3d2 is divisible by 4, then dividing it by 4 results in a quotient also in the form of a2 + 3b2.
023	In order for c2 + 3d2 to be divisible by 4, it must be even, and c and d must have the same parity (either both or even or both or odd).
024	Case 1: both c and d are even:
025	Let c = 2C and d = 2D.
026	Then c2 + 3d2 = (2C)2 + 3(2D)2  = 4C2 + 12D2 = 4(C2 + 3D2).
027	So dividing c2 + 3d2 by 4 results in a quotient also in the form of a2 + 3b2.
028	Case 2: both c and d are odd:
029	Let c = 4C ± 1 and d = 4D ± 1
030	Let j = ±1 and k = ±1, so c = 4C + j and d = 4D + k.
031	Then c2 + 3d2 = (4C + j)2 + 3(4D + k)2 = 16C2 + 8Cj + j2 + 3(16D2 + 8Dk + k2) = 16C2 + 8Cj + j2 + 48D2 + 24Dk + 3k2 = 16C2 + 8Cj + 48D2 + 24Dk + 3k2 + j2 = 16C2 + 8Cj + 48D2 + 24Dk + 4 = 4(4C2 + 2Cj + 12D2 + 6Dk + 1) = 4(C2 + 9D2 + 1 + 6cdjk + 2Cj + 6Dk + 3C2 – 6cdjk + 3D2) = 4(C2j2 + 9D2k2 + 1 + 6cdjk + 2Cj + 6Dk + 3C2j2 – 6cdjk + 3D2k2) = 4((Cj + 3Dk + 1)2 + (Cj – Dk)2) = 4(A2 + 3B2).
032	So dividing c2 + 3d2 by 4 results in a quotient also in the form of a2 + 3b2.
033	Either way, dividing c2 + 3d2 by 4 results in a quotient also in the form of a2 + 3b2.
034	Dividing an integer in the form of c2 + 3d2 by a prime in the form of a2 + 3b2 also results in an integer in the form of a2 + 3b2, namely r2 + 3s2 where r = |j1ac – 3j1j2j3bd| / (a2 + 3b2) and s = |j3ad + j2bc| / (a2 + 3b2) where j1 = ±1, j2 = ±1, and j3 = ±1, and j1, j2, and j3 can be chosen so that r and s are positive integers.
035	r2 + 3s2 = (|j1ac – 3j1j2j3bd| / (a2 + 3b2))2 + 3(|j3ad + j2bc| / (a2 + 3b2))2 = [|j1ac – 3j1j2j3bd|2 + 3|j3ad + j2bc|2] / (a2 + 3b2)2 = [j12a2c2 – 6j12j2j3abcd + 9j12j22j32b2d2 + 3(j32a2d2 + j2j3abcd + j22b2c2)]/ (a2 + 3b2)2 = [a2c2 + 9b2d2 + 3a2d2 + 3b2c2]/(a2 + 3b2)2 = [a2c2 + 3a2d2 + 3b2c2 + 9b2d2]/(a2 + 3b2)2 = [a2(c2 + 3d2) + 3b2(c2 + 3d2)]/(a2 + 3b2)2 = [(a2 + 3b2)(c2 + 3d2)]/(a2 + 3b2)2 = (c2 + 3d2)/(a2 + 3b2)
036	Now (ad – bc)(ad + bc) is divisible by the prime a2 + 3b2 because (ad – bc)(ad + bc) = a2d2 – b2c2 = a2d2 + 3b2d2 – b2c2 – 3b2d2 = d2(a2 + 3b2) – b2(c2 + 3d2) = d2(a2 + 3b2) – b2(a2 + 3b2)(r2 + 3s2) = (a2 + 3b2)[d2 – b2(r2 + 3s2)].
037	Since (ad – bc)(ad + bc) is divisible by the prime a2 + 3b2, then either ad – bc is divisible by a2 + 3b2 or ad + bc is divisible by a2 + 3b2.
038	Case 1: ad – bc is divisible by a2 + 3b2:
039	Then ad – bc = (a2 + 3b2)F
040	And (ac + 3bd) is divisible by the prime a2 + 3b2 because (ac + 3bd)2 = a2c2 + 6abcd + 9b2d2 = a2c2 + 6abcd + 9b2d2 + 3a2d2 – 3a2d2 + 3b2c2 – 3b2c2 = a2c2 + 3b2c2 + 3a2d2 + 9b2d2 – 3a2d2 + 6abcd – 3b2c2 = c2(a2 + 3b2) + 3d2(a2 + 3b2) – 3(a2d2 – 2abcd + b2c2) = c2(a2 + 3b2) + 3d2(a2 + 3b2) – 3(ad – bc)2 = c2(a2 + 3b2) + 3d2(a2 + 3b2) – 3((a2 + 3b2)F)2 = (a2 + 3b2)[c2 + 3d2 – 3(a2 + 3b2)F2].
041	Case 2: ad + bc is divisible by a2 + 3b2:
042	Then ad + bc = (a2 + 3b2)F
043	And (ac – 3bd) is divisible by the prime a2 + 3b2 because (ac – 3bd)2 = a2c2 – 6abcd + 9b2d2 = a2c2 – 6abcd + 9b2d2 + 3a2d2 – 3a2d2 + 3b2c2 – 3b2c2 = a2c2 + 3b2c2 + 3a2d2 + 9b2d2 – 3a2d2 – 6abcd – 3b2c2 = c2(a2 + 3b2) + 3d2(a2 + 3b2) – 3(a2d2 + 2abcd + b2c2) = c2(a2 + 3b2) + 3d2(a2 + 3b2) – 3(ad + bc)2 = c2(a2 + 3b2) + 3d2(a2 + 3b2) – 3((a2 + 3b2)F)2 = (a2 + 3b2)[c2 + 3d2 – 3(a2 + 3b2)F2].
044	In other words, if ad + bc is divisible by a2 + 3b2 then ac – 3bd is divisible by a2 + 3b2, or if ad – bc is divisible by a2 + 3b2 then ac + 3bd is divisible by a2 + 3b2.
045	Since r = |j1ac – 3j1j2j3bd| / (a2 + 3b2) and since s = |j3ad + j2bc| / (a2 + 3b2), values for j1 = ±1, j2 = ±1, and j3 = ±1 can be chosen so that a2 + 3b2 divides positively into |j1ac – 3j1j2j3bd|, and so that a2 + 3b2 divides positively into |j3ad + j2bc|, making both r and s positive integers.
046	If ad – bc and ac + 3bd is divisible by a2 + 3b2, then j1 = 1, j2 = 1, and j3 = -1 or j1 = 1, j2 = -1, and j3 = 1 or j1 = -1, j2 = 1, and j3 = -1 or j1 = -1, j2 = -1, and j3 = 1; or j1j2j3 = 1.
047	If ad + bc and ac – 3bd is divisible by a2 + 3b2, then j1 = 1, j2 = 1, and j3 = 1 or j1 = 1, j2 = -1, and j3 = -1 or j1 = -1, j2 = 1, and j3 = 1 or j1 = -1, j2 = -1, and j3 = -1; or j1j2j3 = -1.
048	Therefore, dividing c2 + 3d2 by y (which is comprised of all odd factors in the form a2 + 3b2 and possibly the factor 4), results in a quotient in the form of a2 + 3b2 that is equal to x.
049	But this contradicts our assumption that x is not in the form of a2 + 3b2, so y must have a positive odd factor x’ not in the form of a2 + 3b2.
050	There must exist a positive odd integer x’ that is a factor of y and that is not in the form of a2 + 3b2.
051	x’ ≠ 1 because 1 is in the form of a2 + 3b2 (1 = 12 + 3(0)2), so 1 < x’.
052	Since x’ divides into y, x’ < y
053	Also recall from above that y < x.
054	So 1 < x’ < y < x which means 1 < x’ < x.
055	Therefore, if x is odd and not in the form of a2 + 3b2, there must exist another odd integer x’ such that 1 < x’ < x that is not in the form of a2 + 3b2.
056	But we can use the same argument to prove the existence of another odd integer x” such that 1 < x” < x’ that is not in the form of a2 + 3b2, and another x’”, and so on, so we have a contradiction by the method of infinite descent.
057	So if a and b are relatively prime, any odd factor x of a2 + 3b2 must also have the same form.

Q.E.D.