Wednesday, January 18, 2006

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

2 comments:

Kevin said...

there seems to be a typo in line (16): it says ζ2^2 where i think it should be ζ^2, correct me if i'm wrong

but great job on the proof(s), they are very clear and helped me a lot with my thesis.

greetings from Belgium

Kevin

Larry Freeman said...

Hi Kevin,

Thanks for noticing the typo. I just fixed it.

I'm very glad that proofs are helpful.

Cheers,

-Larry