Fermat's Last Theorem: Proof for n = 7: x + y + z = 0

In today's blog, I will continue the proof for Fermat's Last Theorem n=7. Today's proof is used in step four of the FLT n=7 proof.

The details for today are based on Paulo Ribenboim's Fermat's Last Theorem for Amateurs.

Lemma 1: (x + y)⁷ - x⁷ - y⁷ = 7xy(x+y)(x² + xy + y²)²

(1) With help from the binomial theorem, we get:

(x + y)⁷ - x⁷ - y⁷ = 7x⁶y + 21x⁵y² + 35x⁴y³ + 35x³y⁴ + 21x²y⁵ + 7xy⁶

(2) (x² + xy + y²)² = x⁴ + 2x³y + 3x²y² + 2xy³ + y⁴

(3) (x + y)(x² + xy + y²)² =
x⁵ + 3x⁴y + 5x³y² + 5x²y² + 3xy⁴ + y⁵

(4) 7xy(x + y)(x² + xy + y²)² =
7x⁶y + 21x⁵y² + 35x⁴y³ + 35x³y⁴ + 7xy⁶

QED

Lemma 2: if ζ =(-1 + √-3)/2, then ζ² = (-1 - √-3)/2

[(-1 + √-3)/2][(-1 + √-3)/2] = (1 - 2√-3 + -3)/4 = (-2 - 2√-3) /4 = (-1 -√-3)/2

QED

Lemma 3: if x⁷ + y⁷ + z⁷ = 0, x,y,z integers, xyz ≠ 0 and x + y + z = 0, then x,y,z are proportional to 1, ζ, ζ² where ζ = (-1 + √-3)/2

(1) x + y + z = 0 → z = -x -y = -(x + y)

(2) x⁷ + y⁷ + z⁷ = 0 → z⁷ = -(x⁷ + y⁷)

(3) (x + y)⁷ - x⁷ - y⁷ = (-z)⁷ + z⁷ = 0

(4) Applying Lemma 1 above, we get:
7xy(x + y)(x² + xy + y²)² = (x + y)⁷ - x⁷ - y⁷ = 0

(5) Since xyz ≠ 0, we know that x ≠ 0 and y ≠ 0 and z ≠ 0.

(6) Since z = -(x + y) and z ≠ 0 we know that x + y ≠ 0.

(7) Therefore x² + xy + y² = 0.

(8) But then dividing all sides by x² gives us:
1 + y/x + (y/x)² = 0

(9) Using the quadratic equation (see here), we get:
y/x = (-1 ± √1 - 4(1)(1))/2 = (-1 ± √-3)/2

(10) This means that y/x = ζ or y/x = ζ²

(11) Which means that y is proportional to ζ or ζ²

(12) z = -(x + y) = -x(1 + y/x)

So that z = -x(1 + ζ) or z=-x(1 + ζ²)

(13) But 1 + ζ = 1 + (-1 + √-3)/2= (2 - 1 +√-3)/2 = (1 + √-3)/2

So that:
(-x)(1 + ζ) = (-x)[(1 + √-3)/2 ] = x[(-1 - √-3)/2 ] = xζ²

(14) And 1 + ζ² = 1 + (-1 -√-3)/2 = (2-1 -√-3)/2 = (1 - √-3)/2

So that:
(-x)(1 + ζ²) = (-x)[(1 - √-3)/2] = x[(-1 + √-3)/2] = xζ

(15) This proves that z is proportional to ζ or ζ²

(16) Finally x/y = 1/ζ or x/y=1/ζ²

(a) 1/ζ = 2/(-1 + √-3) = 2(-1 - √-3)/[1 - (-3)] = (-2 - 2√-3)/4 = (-1 - √-3)/2 = ζ²

(b) 1/ζ² = 2/(-1 -√-3) = 2(-1 + √-3)/[1 - (-3)] = (-2 + 2√-3)/4 = (-1 + √-3)/2 = 1/ζ

(17) So, we see that x too is proportional to either ζ or ζ²

QED