Saturday, January 21, 2006

Fermat's Last Theorem: Proof for n=7: Existence of u

In today's blog, I will continue the proof for Fermat's Last Theorem n=7.

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

Lemma 1: if ax2 + bx + c = 0 where a,b,c are rational numbers and x is a rational number, then b2 - 4ac is the square of a rational number.

(1) We know that x = (-b ±√b2 - 4ac)/2 from the quadratic equation formula (see here).

(2) We know that 2ax is rational since 2 is an integer and a,x are rational.

(3) Let d = √b2 - 4ac

(4) From #1 and #2, we know that -b ± d must also be rational since it is equal to 2ax.

(5) Then d must be rational since:

d = ±(2ax + b) where 2ax is a rational and b is an integer.

QED

Lemma 2: if Q = (t + s)/2t, then ([1 - Q + Q2]/2)2 = ([3t2 + s2]/8t2)2

Proof:

(1) (1 - Q + Q2)/2 = (1 - [(t + s)/2t] + [(t + s)/2t]2)/2 =
= [3t2 + 5s2]/8t2

(2) From this the conclusion follows.

QED

Lemma 3: Given M,Q are rational numbers, M2 - M(1 - Q + Q2) + 1/7 = 0, then there exists a rational number u such that:
u2 = s4 + 6t2s2 - t4/7
s,t are integers
s/t = 2Q-1
gcd(s,t) = 1
t is greater than 0

Proof:

(1) Since M, Q are rational and M2 - M(1 - Q + Q2) + 1/7 = 0, using lemma 1 above, we can conclude that:
[(1 - Q + Q2)/2]2 - 1/7 is the square of a rational number.

(2) Let s/t = 2Q - 1

(a) Since Q is a rational number, we know that there exists s',t' such that Q = s'/t'

(b) 2Q - 1 is therefore also a rational number since:

2Q - 1 = (2s' - t')/t'

(3) We can assume gcd(s,t)=1 since if there were any common divisors, we could divide them out an s/t would still = 2Q - 1.

(4) We know that t ≠ 0 and we can assume that it is positive since we can multiply -1 to s and still have s/t = 2Q-1.

(5) Q = (t + s)/2t since:

2Q = s/t + 1 = (s + t)/t

Q = (s + t)/2t

(6) Using Lemma 2, we note that:
(1 - Q + Q2)2 - 4/7 = 4[(1 - Q + Q2)2 - 1/7] = 4[(3t2 + s2)2/(8t2)2 - 1/7]

(7) Applying #1 with #6, we can conclude that:
(3t2 + s2)2/(8t2)2 - 1/7 is the square of a rational number.

(8) Let u2 = 64t4[(3t2 + s2)2/(8t2)2 - 1/7]

(9) Since 64t4 is the square of a rational number and using #7, we can conclude that u2 is the square of a rational number which means that u is a rational number.

(10) Further, we note that:
64t4[(3t2 + s2)2/(8t2)2 - 1/7] = s4 + 6t2s2 - t4/7.

QED

Scouse Rob said...

In Lemma 1 step (1) should:

x = (-b ±√b^2 - 4ac)/2

be

x = (-b ±√b^2 - 4ac)/2a

Rob

Scouse Rob said...

In Lemma 2 step (1) should:

(1 - [(t + s)/2t] + [(t + s)/2t]^2)/2
= [3t2 + 5s^2]/8t^2

be

(1 - [(t + s)/2t] + [(t + s)/2t]^2)/2
= [3t2 + s^2]/8t^2

Rob