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
Wednesday, January 18, 2006
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2 comments:
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
Hi Kevin,
Thanks for noticing the typo. I just fixed it.
I'm very glad that proofs are helpful.
Cheers,
-Larry
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