## Sunday, January 15, 2006

### Fermat's Last Theorem: Proof for n=5: Key Lemma 1

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,x such that:

(a) a,b,x are integers
(b) a,b have different parities
(c) gcd(a,b)=1
(d) a,b are nonzero
(e) 5 doesn't divide a
(f) 5 divides b
(g) a2 - 5b2 = x5
(h) there exist integers h,k such that: a + b√5 = ([h + k√5]/2)5
(i) h,k have the same parity

Then h,k are even.

(1) 25b = 5k(h4 + 10h2k2 + 5k4) since:

(a) (h + k√5)5 = h5 + 5h4k√5 + 50h3k2 + 50h2k35 + 125hk4 + 25k55.

(b) From (a) and from (h) above: 25b = 5h4k + 50h2k3 + 25k5 [25 is necessary since a + b√5 = [h + k√5] divided by 25

(c) From (b), 25b = 5k(h4 + 10h2k2 + 5k4)

(2) Assume that h,k are odd

(3) From (c), we know that 32 must divide (h4 + 10h2k2 + 5k4) since 2 doesn't divide 5 and since 2 doesn't divide k (since we are assuming in #2 that k is odd)

(4) h4 or k4 ≡ 1 or 17 (mod 32) since (the argument for h below applies equally to k):

(a) h is odd so h ≡ 1 (mod 2) which means that h modulo 32 could be any odd number including ±1, ± 3, ± 5, and so on until ± 15.

(b) h2 modulo 32 can only be congruent to 1, 9, 17, or 25 (see here for a review of modular arithmetic)

(±1)2 ≡ 1 (mod 32)
(±3)2 ≡ 9 (mod 32)
(±5)2 ≡ 25 (mod 32)
(±7)2 ≡ 49 ≡ 17 (mod 32)
(±9)2 ≡ 81 ≡ 17 (mod 32)
(±11)2 ≡ 121 ≡ 25 (mod 32)
(±13)2 ≡ 169 ≡ 9 (mod 32)
(±15)2 ≡ 225 ≡ 1 (mod 32)

(c) h4 modulo 32 can only be congruent 1 or 17.

(1)2 ≡ 1 (mod 32)
(9)2 ≡ 17 (mod 32)
(17)2 ≡ 289 ≡ 1 (mod 32)
(25)2 ≡ 625 ≡ 17 (mod 32)

(5) For h4 ≡ 1 (mod 32), we know that h2 ≡ 1 or 17 (mod 32)

(6) For h4 ≡ 17 (mod 32), we know that h2 ≡ 9 or 25.

(7) But from our assumption in #2, we find that 32 cannot divide (h4 + 10h2k2 + 5k4) since modulo 32 it is congruent to 16 (see below)

Case h4 ≡ 1, k4 ≡ 1:

(1) + 10(1)(1) + 5(1) ≡ 1 + 10 + 5 ≡ 16 (mod 32)
(1) + 10(17)(1) + 5(1) ≡ 1 + 170 + 5 ≡ 1 + 10 + 5 ≡ 16 (mod 32)
(1) + 10(1)(17) + 5(1) ≡ 1 + 170 + 5 ≡ 1 + 10 + 5 ≡ 16 (mod 32)
(1) + 10(17)(17) + 5(1) ≡ 1 + 2890 + 5 ≡ 1 + 10 + 5 ≡ 16 (mod 32)

Case h4 ≡ 17, k4 ≡ 1:

(17) + 10(9)(1) + 5(1) ≡ 17 + 90 + 5 ≡ 17 + 26 + 5 ≡ 16 (mod 32)
(17) + 10(25)(1) + 5(1) ≡ 17 + 250 + 5 &eqiuv; 17 + 26 + 5 ≡ 16 (mod 32)
(17) + 10(9)(17) + 5(1) ≡ 17 + 1530 + 5 ≡ 17 + 26 + 5 ≡ 16 (mod 32)
(17) + 10(25)(17) + 5(1) ≡ 17 + 4250 + 5 ≡ 17 + 26 + 5 ≡ 16 (mod 32)

Case h4 ≡ 1, k4 ≡ 17:

(1) + 10(1)(9) + 5(17) ≡ 1 + 90 + 85 ≡ 1 + 26 + 21 & equiv; 16 (mod 32)
(1) + 10(17)(9) + 5(17) ≡ 1 + 1530 + 85 ≡ 1 + 26 + 21 & equiv; 16 (mod 32)
(1) + 10(1)(25) + 5(17) ≡ 1 + 250 + 85 &eqiuv; 1 + 26 + 21 ≡ 16 (mod 32)
(1) + 10(17)(25) + 5(17) ≡ 1 + 4250 + 85 ≡ 1 + 26 + 21 & equiv; 16 (mod 32)

Case h4 ≡ 17, k4 equiv; 17

(17) + 10(9)(9) + 5(17) ≡ 17 + 810 + 85 ≡ 17 + 10 + 21 ≡ 16 (mod 32)
(17) + 10(25)(9) + 5(17) ≡ 17 + 2250 + 85 ≡ 17 + 10 + 21 ≡ 16 (mod 32)
(17) + 10(9)(25) + 5(17) ≡ 17 + 2250 + 85 ≡ 17 + 10 + 21 ≡ 16 (mod 32)
(17) + 10(25)(25) + 5(17) ≡ 17 + 6250 + 85 ≡ 17 + 10 + 21 ≡ 16 (mod 32)

(8) So we have a contradiction and we reject our assumption.

QED

Lemma 2: Given a,b such that:

(a) gcd(a,b)=1
(b) a,b have different parities (one is odd, one is even)
(c) a,b are nonzero
(d) 5 doesn't divide a
(e) 5 divides b
(f) a + b√5 = [(m + n√5)/2]5([t+u√5]/2) where u = 0.

Then:

(a) a = c(c4 + 50c2d2 + 125d4)
(b) b = 5d(c4 + 10c2d2 + 5d4)
(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 there exists m',n' such that:
(m' + n'√5)/2 = [(m+n√5)/2]5= (m5 + 5m4n√5 + 50m3n2 + 50m2n35 + 125mn4 + 25n55)/25

(2) So, m' = (m5 + 50m3n2 + 125mn4)/24

(3) This means that 16m' ≡ m5 (mod 5)

(4) So, n' = (5m4n + 50m2n3 + 25n5)/24

(5) This means that 16n' ≡ 0 (mod 5) and 5 divides n'.

(6) a + b√5 = [(m' + n'√5)/2][(t + u√5)/2] = (m't + m'u√5 + n't√5 + 5n'u)/4

(7) So:
4a = m't + 5n'u
4b = m'u + n't

(8) From this we can conclude that 5 doesn't divide m' (otherwise from #7, 5 would divide a which it does not)

(9) We can also conclude that 5 doesn't divide m (otherwise from #3, 5 woud divide m' which from #8 it does not)

(10) From #7, we know that 5 divides u (since it divides 4b, n't but doesn't divide m')

(11) We also know that t = ± 2 since u = 0 and (t + u√5)/2 is a unit.

(12) From #11, we know that a + b√5 = ± [(m + n√5)/2]5

(13) From #12 and Lemma 1 above and the fact that m ≡ n (mod 2), we can conclude that m,n are even.

(14) Let:
c = ±m/2
d = ±n/2

(15) a + b√5 = ±(c + d√5)5 = c5 + 5c4d√5 + 50c3d2 + 50c2d35 + 125cd4 + 25d55.

(16) So:
a = c5 + 50c3c2 + 125cd4 = c(c4 + 50c2d2 + 125d4)

b = 5c4d + 50c2d3 + 25d5 = 5d(c4 + 10c2d2 + 5d4)

(17) gcd(c,d) = 1 since c divides a, d divides b. If gcd(c,d) ≠ 1 then it would mean a,b have a common factor which they don't.

(18) c,d have different parities

(a) c,d cannot both be even since gcd(c,d)=1 from #17

(b) c,d cannot both be odd since from #16:

a = odd(odd + even + odd) = odd(even) = even

b = odd(odd + even + odd) = odd(even) = even

Which is impossible since a,b have different parities.

(c) Therefore c,d have different parities too.

(19) 5 doesn't divide c since 5 doesn't divide m from #9.

(20) Finally, we know that c,d are nonzero since:

(a) c is nonzero from #16 since a is nonzero

(b) d is nonzero from #16 since b is nonzero

QED

Lemma 3: Given a,b such that:

(a) gcd(a,b)=1
(b) a,b have different parities (one is odd, one is even)
(c) a,b are nonzero
(d) 5 doesn't divide a
(e) 5 divides b
(f) a + b√5 = [(m + n√5)/2]5([t+u√5]/2) where u ≠ 0.

Then:

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

Proof:

(1) Since [t+u√5]/2 is a unit, we know that there exists an integer e (see here) such that:

[t+u√5]/2 = ±([1 + √5]/2)e

(2) By our assumption that u ≠ 0, we know that e ≠ 0

(3) We can assume that e is positive, since we make e positive by replacing [1 + √5]/2 with its inverse -(1 - √5)/2 since:

(a) [(1 + √5)/2]-1 = 2/(1 + √5) = 2(1 - √5)/(1 - 5) = (2 - 2√5)/(-4) = (1 - √5)/(-2) = -(1 - √5)/2

(b) [(1 + √5)/2]-e = ([(1 + √5)/2]-1)e = [-(1 - √5)/2 ]e

(4) We know that e is ≥ 2 since:

(a) Since 5 divides b, it must divide u (see the reasoning in lemma 2)

(b) But if e = 1, then u = ±1 which is not divisible by 5.

(5) So, we can assume that (t + u√5)/2 = [±(1  √5)e]/(2e)

(6) Which implies that:
±(2e-1)(t + u√5) = (1 ±  √5)e

(7) Using the binomial theorem (see here), we see that 2e-1u ≡ ±e (mod 5) since:

(1 ± √5)e = 1 + e√5 + [e*(e-1)]5 + [e*(e-1)(e-2)](5)√5 + ... + (√5)e

Since e ≥ 2, we know that all other value but the second one are divisible by 5.

(8) Since 5 divides u (from #4a), we know that e ≡ 0 (mod 5) so that 5 divides e.

(9) Then there must exist f such that e = 5f.

(10) We can assume that there exist c',d' such that:
(c' + d'√5)/2 = [(m + n√5)/2][(1 ± √5)/2]f

(11) From this we know that:
a + b√5 = [(c' + d'√5)/2]

(12) We further know that c', d' have the same parity (because of the nature of Z[(1 + √5)/2]) so that we can assume that they are even from Lemma 1 above.

(13) This allows us to suppose there c,d such that:
c = ±c'/2, d = ±d'/2

(14) We now have the equation:
a + b√5 = ±(c + d√5)5

(15) This allows to follow the same logic from Lemma 2, starting at step #15.

QED

Scouse Rob said...

In Lemma 1 step (1b) should

2^5 is necessary since a + b√5 = [h + k√5] divided by 2^5

be

2^5 is necessary since a + b√5 = [h + k√5]^5 divided by 2^5

Scouse Rob said...

In Lemma 2 step (16) should

a = c^5 + 50c^3c^2 + 125cd^4

be

a = c^5 + 50c^3d^2 + 125cd^4

Scouse Rob said...

In the statements and proofs of Lemma 2 and Lemma 3 shouldn't we also have to prove:

5 does divide d

from

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

Rob

Scouse Rob said...

In Lemma 3 step (11) should

a + b√5 = [(c' + d'√5)/2]

be

a + b√5 = [(c' + d'√5)/2]^5

Rob