Fermat's Last Theorem: Proof for n = 5: Key Lemma 2

Today's blog continues the proof for Fermat's Last Theorem: n= 5. If you are interested in the history behind the proof, start here. If you are interested in the mathematical details behind the proof, start here.

Today's blog rests on the proof offered by Paulo Ribenboim in Fermat's Last Theorem for Amateurs.

Lemma 1: Given a,b such that:

(a) gcd(a,b)=1
(b) a ≡ b (mod 2)
(c) a,b are nonzero
(d) 5 doesn't divide a
(e) 5 divides b
(f) (a + b√5)/2 = [(m + n√5)/2]⁵([t+u√5]/2) where u = 0.

Then:

(a) a = c(c⁴ + 50c²d² + 125d⁴)/16
(b) b = 5d(c⁴ + 10c²d² + 5d⁴)/16
(c) gcd(c,d)=1
(d) c,d have different parities
(e) 5 doesn't divide c
(f) c,d are nonzero

Proof:

(1) t² - 5u² = ± 4 since [t+u√5]/2 is a unit.

(2) Since u = 0, t = ± 2 so t/2 = ± 1.

(3) Let c = ±m, d = ± n

(4) We now have:
(a + b√5)/2 = ±[(c + d√5)/2]⁵.

(5) a + b√5 = (c⁵ + 5c⁴d√5 + 50c³d² + 50c²d³√5 + 125cd⁴ + 25d⁵√5)/16

(6) So that:

a = (c⁵ + 50c³d² + 125cd⁴)/16 = c(c⁴ + 50c²d² + 125d⁴)/16

b = (5c⁴d + 50c²d³ + 25d⁵)/16 = 5d(c⁴ + 10c²d¹⁰ + 5d⁴)/16

(7) gcd(c,d)=1 since c divides a, d divides b, and gcd(a,b) = 1.

(8) c,d are odd since a,b are both odd (since c divides a and d divides b)

(9) 5 doesn't divide c since 5 doesn't divide a.

QED

Lemma 2: Given a,b such that:

(a) gcd(a,b)=1
(b) a ≡ b (mod 2)
(c) a,b are nonzero
(d) 5 doesn't divide a
(e) 5 divides b
(f) (a + b√5)/2 = [(m + n√5)/2]⁵([t+u√5]/2) where u ≠ 0.

Then:

(a) a = c(c⁴ + 50c²d² + 125d⁴)/16
(b) b = 5d(c⁴ + 10c²d² + 5d⁴)/16
(c) gcd(c,d)=1
(d) c,d have different parities
(e) 5 doesn't divide c
(f) c,d are nonzero

Proof:

(1) We know that t² - 5u² = ± 4 since it is a unit.

(2) We know that we can assume that;
(t + u√5)/2 = ±[(1 ± √5 )/2]^e where e is ≥ 2 and 5 divides e (see the reasoning in Lemma 3 here)

(3) This means that we can define c,d such that:
(c + d√5)/2 = ±([m + n√5]/2)([1 ± √5]/2)^f

(4) This gives us:
(a + b√5)/2 = [(c + d√5)/2]⁵

(5) We can then follow the logic from step #4 in Lemma 1 above.

QED