Fermat's Last Theorem: n = 3: a2 + 3b2

The essence of Euler's proof for Fermat's Last Theorem, n = 3, is realizing the properties of values that can be represented by the formula: a² + 3b².

In today's blog, I will present a lemma that illustrates one of the surprising properties of the above formula.

This blog will continue the details of the proof that was begun in a previous blog.

Lemma: If gcd(a,b)=1, then every odd factor of a² + 3b² has this same form.

In other words, for every odd factor of a² + 3b², there exists c,d such that the odd factor = c² +3d²

(1) Let x be a positive, odd factor of a² + 3b²

(2) Then, there exists a value f such that:
a² + 3b² = xf

(3) We can assume that x is greater than 1 (since if x = 1 it's only factor is itself and 1 = 1² + 3(0)².)

(4) There exists integers m,n such that:

a = mx ± c
b = nx ± d

where c,d can be positive or negative.

(5) Now, we can assume that c,d have absolute values that are less than (1/2)x.

(a) From the Division Algorithm (see Theorem 1, here), we know that c is less than x.

(b) Assume that c ≥ (1/2)x

(c) c' = x - c = -(c - x)

(d) a = mx + c = (m+1)x + c - x = (m+1)x - (x - c) = (m+1)x - c'

(e) abs(c') is less than (1/2)x since c' = x - c

(6) From step #4, we have:

a² + 3b² =

(mx ± c)² + 3(nx ± d)² =

m²x² ± 2mxc + c² + 3n²x² ± 6nxd + 3d² =

= x(m²x ± 2mc + 3n²x ± 6nd) + c² + 3d²

(7) So x divides c² + 3d² since from step #1, since x is a factor of a² + 3b².

(8) So, there exists y such that:

c² + 3d² = xy

(9) Now from step #5 above, we have:

xy = c² + 3d² is less than [(1/2)x]² + 3[(1/2)x]² = (1/4)x² + (3/4)x² = x².

(10) Now, we can conclude that y is less than x from:

(a) Since c², d² are nonnegative, it follows that xy is nonnegative.

(b) From step #9, xy is less than x²

(c) Since x is positive, the only way that this can be true is if y is less than x.

(11) We also know that c² + 3d² ≠ 0 because:

(a) Assume c² + 3d² = 0.

(b) Then c=0, d=0 since c² and 3d² are both nonnegative.

(c) But then: a = mx, b = nx and x divides both a,b.

(d) Which contradicts gcd(a,b)=1 so we reject our assumption.

(12) Let g = gcd(c,d). We know there exists C,D such that c = gC, d=gD, gcd(C,D)=1. [From the properties of greatest common denominators]

(13) Now we also know that g² divides y because:

(a) xy =c² +3d² = (gC)² + 3(gD)² = g²(C² + 3D²)

(b) Assume there is a prime p that divides g but doesn't divide y.

(c) Then p divides g and p divides x

(d) Then, there exists X,G where x = Xp, g = Gp

(e) And, p divides a and b since

a = mx + c = mpX + GpC = p(mX + GC)

b = nx + d = npX + GpD = p(nX + GD)

(f) But this is impossible since a,b are coprime.

(g) So, there is no prime p which divides g but does not divide y.

(h) Which proves that g divide y and further that g² divides y

(14) Since g² divides y, there exists z such that:

y = g²z

(15) Now, it follows that there exist C,D such that:

xz = C² + 3D² where gcd(C,D)=1

(a) xy = g²(C² + 3D²) [from step #13a above]

(b) gcd(C,D) = 1 [from step #12 above]

(c) Then substituting z from step #14 gives us:

xz = C² + 3D²

(16) Now we have enough to conclude that x is of the form p² + 3q²

(a) Assume that x is not of this form.

(b) From step #15, we have xz = C² + 3D² where gcd(C,D)=1.

(c) Then there must exist w such that w divides z and w is not of the form p² + 3q² [See here
for the proof of this lemma]

(d) Now, w ≠ 1 since 1 = 1² + 3(0)²

(e) So, w is smaller than z [since w is greater than 1 and w divides z]

(f) But then w is less than x [since z is less than y (step #14, since y = g²z) and y is less than x (step #10)]

(g) But now, we have proven that the existence of x proves the existence of a smaller factor w which divides a smaller value of the same form.

(h) We can use the same argument to presuppose a value w' which is smaller than w and another value w'' which is smaller than w'. In other words, we have a contradiction by the method of infinite descent.

QED