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