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

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 just interested in the mathematical details behind the proof, start here.

Today's blog continues the proof on Fermat's Last Theorem: n=5 by offering two lemmas that are really the crucial idea behind the proof. Both lemmas rest on foundations of continued fractions and quadratic integers.

Lemma 1:

Given two integers a,b such that:

(a) gcd(a,b)=1
(b) a,b have different parities (one is odd, one is even)
(c) 5 doesn't divide a
(d) 5 divides b
(e) (a² - 5b²) is a fifth power.

Then there exists two integers c,d such that:

(i) a = c(c⁴ + 50c²d² + 125d⁴)
(ii) b = 5d(c⁴ + 10c²d² + 5d⁴)
(iii) gcd(c,d)=1
(iv) c,d have different parities
(v) 5 doesn't divide c
(vi) 5 does divide d
(vii) c,d are nonzero

Proof:

(1) Assume that gcd(a + b√5, a - b√5)= d which is greater than 1.

(2) Then there exists e,f such that:
de = a + b√5
df = a - b√5

(3) a + b√5 + a - b√5 = 2a = de + df = d(e + f)

(4) Likewise a + b√5 - (a - b√5) = 2b√5 = d(e - f)

(5) Now d doesn't divide √5 since:

(a) √5 is a prime (see here).

(b) So if d divides it, it would have to = √5

(c) But d divides 2a (from #3)

(d) Which would mean √5 divides 2a which means 5 divides 4a²

(e) Which is impossible since 5 doesn't divide a.

(6) So d divides 2b.

(7) So d must divide 2 since it divides 2a and 2b but gcd(a,b)=1.

(8) Now 2 is prime so d=2. (See here)

(9) This means that 4 divides (a + b√5)(a - b√5) = a² - 5b²

(10) But this is impossible since a,b have different parities:
(odd) - 5(even) = odd - even = odd
(even) - 5(odd) = even - odd = odd

So we reject our assumption.

(11) Since Z[(1 + √5)/2] has unique factorization (see here), we can conclude that a + b√5 and a - b√5 are fifth powers (see here ).

(12) This means that there exists m,n,t,u such that a + b√5 = [(m + n√5)/2]⁵[(t + u√5)/2] where (t + u√5)/2 is a unit.

(13) Now, if u = 0, then we reach our conclusion (see here). If u ≠ 0, we reach our conclusion (see here )

QED

Lemma 2:

Given two integers a,b such that:

(a) gcd(a,b)=1
(b) a,b are both odd
(c) 5 doesn't divide a
(d) 5 divides b
(e) (a² - 5b²)/4 is a fifth power.

Then there exists two integers c,d such that:

(i) a = c(c⁴ + 50c²d² + 125d⁴)/16
(ii) b = 5d(c⁴ + 10c²d² + 5d⁴)/16
(iii) gcd(c,d)=1
(iv) c,d are both odd
(v) 5 doesn't divide c
(vi) 5 divides d
(vii) c,d are nonzero.

(1) Assume that gcd([a + b√5]/2, [a - b√5]/2)= d which is greater than 1.

(2) Then there exists e,f such that:
de = (a + b√5)/2
df = (a - b√5)/2

(3) (a + b√5)/2 + (a - b√5)/2= a = de + df = d(e + f)

(4) Likewise (a + b√5)/2 - (a - b√5)/2 = b√5 = d(e - f)

(5) Now d doesn't divide √5 since:

(a) √5 is a prime (see here).

(b) So if d divides it, it would have to = √5

(c) But d divides a (from #3)

(d) Which is impossible since 5 doesn't divide a.

(6) So d divides b.

(7) But this is impossible since gcd(a,b)=1.

So we reject our assumption.

(8) Since Z[(1 + √5)/2] has unique factorization (see here -- to be added later), we can conclude that (a + b√5)/2 and (a - b√5)/2 are fifth powers (see here -- to be added later).

(9) This means that there exists m,n,t,u such that (a + b√5)/2 = [(m + n√5)/2]⁵[(t + u√5)/2] where (t + u√5)/2 is a unit.

(13) Now, if u = 0, then we reach our conclusion (see here ). If u ≠ 0, we reach our conclusion (see here )

QED

To show the proof for these lemmas, it needs to be possible to factor a² - 5b² into (a - b√5)(a + b√5) [See here for details on unique factorization and the existence of √5 as a prime].

To avoid making Euler's mistake, we will need to make sure that Z[√5] or some variant such as Z[(1 + √5)/2] is characterized by unique factorization. More details can be found here.