Fermat's Last Theorem: Proof for n=5: 2 is a prime in Z[(1 + √5)/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 a proof offered by Jody Esmonde and M. Ram Murty in Problems in Algebraic Number Theory.

To understand the details of today's proof, you will need to be familiar with Continued Fractions.

Lemma 1: if d = a² + 1 and a ≥ 1, then the continued fraction expansion of √d is [a , 2a, ... ]

(1) a₀ = a since a is less than √a² + 1 which is less than a + 1 since:

a =√a²
a + 1 = √a² + 2a + 1

a² is less than a² + 1 is less than a² + 2a + 1.

(2) a₂ = 2a since 2a is less than 1/(√a² + 1 - a) which is less than 2a + 1.

(a) 1/(√a² + 1 - a) = (√a² + 1 + a)/(a² + 1 - a²) = √a² + 1 + a.

(b) From (a), we see that 2a is less than a + √a² + 1 since a² + 1 is greater than a².

(c) Since a² + 1 is less than a² + 2a + 1, we see that:

√a² + 1 + a is less than a + (a + 1) = a + √a² + 2a + 1

(d) Finally, we see that all other items from this point a_i are = 2a.

1/(√a² + 1 + a - 2a) = 1/(√a² + 1 - a) =
= (√a² + 1 + a)/(a² + 1 - a²) = √a² + 1 + a.

QED

Lemma 2: if p_k/q_k is a convergent for the continued fraction expansion of α, then gcd(p_k,q_k) = 1.

(1) Let f be a factor that divides both p_k and q_k.

(2) We know that p_kq_k-1 - p_k-1q_k = (-1)^k-1 (see Lemma 2 here)

(3) From (1), we know that f divides p_kq_k-1 - p_k-1q_k which means that f divides (-1)^k-1

(4) But the only way that f divides (-1)^k-1 is if f = 1 since all values are integers.

QED

Lemma 3:
If:
(a) α is an irrational number
(b) r,s are integers
(c) s is greater than 0
(d) absolute(sα - r) is less than absolute(q_kα - p_k) where p_k,q_k are the convergents of the continued fraction expansion of α
Then:
s ≥ q_k+1

Proof:

(1) Assume that 1 ≤ s which is less than q_k+1

(2) Then, there exist x,y such that:
p_kx + p_k+1y = r
q_kx + q_k+1y = s

(3) Multiplying q_k to the first equation and p_k to the second gives us:
p_kq_kx + p_k+1q_ky = rq_k
p_kq_kx + p_kq_k+1y = sp_k

(4) Now, subtracting the second from the first gives us:

y(p_k+1qk - p_kqk+1) = rq_k - sp_k

(5) Multiplying q_k+1 to the first equation in #2 and p_k+1 to the second equation and then subtracting gives us:

x(p_kq_k+1 - p_k+1q_k) = rq_k+1 - sp_k+1

(6) Applying a previous result (see Lemma 2 here), we get:
x = (-1)^k(sp_k+1 - rq_k+1)
y = (-1)^k(rq_k - sp_k)

(7) We know that x ≠ 0

(a) Assume x = 0

(b) r/s = pk_k+1/q_k+1 from #2.

(c) We know that gcd(p_k+1,q_k+1)=1 from Lemma 2 above.

(d) So from (b), r(q_k+1) = s(p_k+1) and from (c), we know that:
q_k+1 must divide s.

(e) But since s is greater than 0, this means that q_k+1 ≤ s which contradicts our assumption at #1 so we can reject (7a).

(8) We know that y ≠ 0

(a) Assume that y = 0

(b) Then r = p_kx

(c) Then s = q_kx

(d) So that absolute(sα - r) = absolute(x) * absolute(q_kα - p_k)

(e) Because x ≠ 0, we then have absolute(x) * absolute(q_kα-p_k) ≥ absolute(q_kα - p_k)

(f) But combining (d) with (e) gives us:
absolute(sα - r) ≥ absolute(q_kα - p_k) which contradicts our original assumption (d) in the lemma statement.

(g) Therefore, we reject (8a).

(9) x,y have opposite signs (one is positive and one is negative)

(a) Assume y is negative then q_kx = s - q_k+1y. Since q_i ≥ 0 (see Corollary 4.1 here) and since s ≥ 1, x is positive.

(b) Assume that y is positive, then q_k+1y ≥ q_k+1 which is greater than s (by assumption #1) so q_kx = s - q_k+1y is less than 0.

(10) We know that α lies in between consecutive convergents since:

(a) By a previous result (see Corollary 5.2 here), we know that if k is even:

p_k/q_k is less than α which is less (p_k+1)/q_k+1

Which means that p_k is less than αq_k and that αq_k+1 is less than p_k+1.

This then gives us that:
q_kα - p_k is positive and q_k+1α - p_k+1 is negative.

(b) We also know that if k is odd,

(p_k+1)/q_k+1 is less than α which is less than p_k/q_k

Which means that p_k+1 is less than αq_k+1 and q_k+1α is less than p_k.

This then gives us that:
q_k+1α - p_k+1 is positive and q_kα - p_k is negative.

(11) By 10(a) and 10(b), we know that regardless of whether k is odd or even, q_kα - p_k and q_k+1α - p_k+1 have opposite signs.

(12) This means that x(q_kα - p_k) and y(q_k+1α - p_k+1) have the same sign.

(13) Finally, this gives us that:

absolute(sα - r) =
= absolute( [q_kx + q_k+1y]α - [p_kx + p_k+1y]) =
= absolute( x[q_kα - p_k] + y[q_k+1α - p_k+1] )

(14) This then gives us:

absolute(sα - r) ≥ absolute(x)*absolute(q_kα - p_k) + absolute(y)*absolute(q_k+1α - p_k+1) ==>
absolute(sα - r) ≥ absolute(x)*absolute(q_kα - p_k) ==>
absolute(sα - r) ≥ absolute(q_kα - p_k)

(15) But this contradicts #1 so we are done.

QED

Lemma 4: 1 ≤ s is less than q_k+1, then absolute(q_kα - p_k) ≤ absolute(sα - r).

(1) Assume that absolute(sα - r) is less than absolute(q_kα - p_k)

(2) Then s ≥ q_k+1

(3) But s is less than q_k+1

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

QED

Lemma 5: if absolute(α - r/s) is less than 1/(2s²) and s is greater than 0, then r/s is a convergent of the continued fraction of α

(1) Assume that r/s is not a convergent of α

(2) Then, r/s ≠ p_n/q_n for all n.

(3) Since q₀ = 1 and q_k ≥ k (see Corollary 4.1 here), we may define a value k such that:

q_k ≤ s which is less than q_k+1

(4) So we also have q_k ≤ s is less than q_k+1

(5) absolute(α - r/s) is less than 1/(2s²) implies that:
absolute(sα - r) is less than 1/2s.

(6) From a result in Lemma 4 above, absolute(q_kα - p_k) ≤ absolute(sα - r).

Which then gives us:
absolute(q_k α - p_k) is less than 1/2s.

(7) Dividing all sides by q_k gives:
absolute(α - p_k/q_k) is less than 1/(2sq_k)

(8) Since r/s ≠ p_k/q_k, absolute(sp_k - rq_k) ≥ 1.

(9) This means that:

1/(sq_k) ≤ absolute(sp_k - rq_k)/sq_k =
= absolute(p_k/q_k - r/s) =
= absolute(p_k/q_k - r/s + α - α )

which is:

≤ absolute(α - p_k/q_k) + absolute(α - r/s)

which is less than:

1/(2sq_k) + 1/2s²

(10) Putting it all together gives us:

1/sq_k is less than 1/(2sq_k) + 1/(2s²)

(11) Subtracting 1/2q_k from both sides gives us:

1/(2sq_k) is less than 1/(2s²).

(12) But this means that:

q_k is greater than s.

(13) This is a contradiction so we reject our assumption #1.

QED

Lemma 6: if x² - dy² is positive and is less than √d, then x,y is a convergent to the continued fraction expansion of √d.

(1) By our assumption above:
x² - dy² = (x + y√d)(x - y√d) is greater than 0.

(2) This means that x > y√d (otherwise, the value in #1 would be negative since x,y are positive numbers)

(3) So x/y is greater than √d

(4) So absolute(x/y - √d) = (x - y√d)/y =

= (x² - dy²)/[y(x + y√d)]

(5) This is less than:

(x² - dy²)/[y(2y√d)] since x is greater than y√d and y(x + y√d) is greater than y(y√d + y√d) = y(2y√d)

(6) And (x² - dy²)/y(2y√d) is less than √d/y(2y√d) [By the assumption of this lemma]

(7) Dividing both sides by √d gives us:

x/y - √d is less than 1/(2y²)

(8) Which by Lemma 5 above gives us our conclusion.

QED

Lemma 7: if x² - dy² is a negative number greater than -√d, then x,y are convergents in the continued fraction expansion of √d.

(1) y² - (1/d)x² is a positive number less than 1/√d since:

(a) √d is greater than dy² - x² (By multiplying by -1 to the assumption)

(b) √d/d is greater than y² - (x²)/d which is greater than 0.

(c) 1/√d is greater than y² -(x²)/d which is greater than 0.

(2) y is greater than x/√d since:

(a) (y - x/√d)(y + x/√d) is greater than 0.

(b) if y were less than this same value would be negative.

(c) if y = x/√d then (2a) would = 0 which it does not.

(3) From #2, y/x is greater than 1/√d.

(4) y/x - 1/√d = (y -x/√d)/x =

= [y² - x²/d]/[x(y + x/√d)]

which is less than:

[y² - x²/d]/(2x²/√d) since:

(a) y is greater than x/√d (from #2 above) so xy is greater than x²/√d.

(b) And xy + x²/√d is greater than 2x²/√d.

(5) Now:
y² - (x²/d) is less than 1/√d (from #1) so:

(y² - x²/√d)/(2x²/√d) is less than: (1/√d)/(2x²/√d)

(6) Since:

(1/√d)/(2x²/√d) = 1/2x², we have the following:
y/x - 1/√d is less than 1/2x².

(7) By the lemma above, y/x is a convergent.

Lemma 8: if d = a² + 1 and absolute(u² - dv²) ≠ 0,1, then absolute(u² - dv²) is greater than √d

(1) The continued fraction for √a² + 1 is [ a, 2a ... ] (see Lemma 1 above)

(2) This means that the period = 1 starting at 1.

(3) From a previous result, this means that for all convergents:
p_k² - dq² = (-1)^k

(4) Now if absolute(u² - dv²) is less than √d, then u,v are convergents since:

(a) If u² - dv² is less than √d, then from Lemma 6 above, u,v are convergents.

(b) If -(u² - dv²) is less than √d, then from Lemma 7 above, u,v are convergents.

(5) From #4, since we are assuming that absolute(u² - dv²) ≠ 1 or 0, then we can conclude that:

absolute(u² - dv²) is greater than √d.

QED

Lemma 9: 2 is a prime in Z[(1 + √5)/2]

(1) Norm(2) = [(4 + 0)/2]*[(4 + 0)/2] = 16/4 = 4

(2) Let's assume that 2 is not a prime.

(3) Then there exists two values a,b that are not units Norm(a) * Norm(b) = Norm(2) = ±4.

(4) Since a,b are not units, we can assume that there norm(a), norm(b) ≠ ± 1.

(5) This limits us to consider Norm(a) = ± 2 and Norm(b) = ± 2.

(6) If Norm(a) = ± 2, then

Norm(a) = [(a + b√5)/2][(a - b√5)/2] = (a² - 5b²)/4

(7) This means that:

(a² - 5b²)/4 = ± 2 which implies that a² - 5b² = ± 8.

(8) We know from properties of Z[(a + b√5)/2] that a,b have the same parity (see here)

(9) First, we note that they cannot be both even.

(a) Assume that u,v are both even

(b) Then there exists u,v such that:
a = 2u
b = 2v

(c) (2u)² - 5(2v)² = ± 8

So:

4u² -20v² = 4(u² - 5v²) = ± 8

And:

u² - 5v² = ± 2.

(d) But absolute(u² - 5v²) ≠ ± 2 from Lemma 8 above since 5 = 2² + 1.

(e) So we reject our assumption and conclude that a,b are both odd.

(10) But they cannot both be odd since:

(a) Assume they are both odd.

(b) Then, there exist u,v such that:
a = 2u + 1
b = 2v + 1

(c) (2u+1)² - 5(2v+1)² = 4u² + 4u + 1 - 5(4v² + 4v + 1) =
= 4u² + 4u + 1 - 20v² -20v - 5 =
= 4u² - 4u - 4 - 20v² - 20v =
= 4(u² + u - 1 - 5v² - 5v) = ± 8

(d) So we can conclude that:
u² + u - 1 -5v² - 5v = ± 2

And:
u(u+1) - 5v(v+1) - 1 = ± 2

(e) But we know that u(u+1) is even and 5v(v+1) is even.

We know they are even since either x or x+1 is even and even * odd = even.

(f) But this means that #10d is impossible since

even - even - odd = odd

(g) So we have again found a contradiction.

(11) But it is impossible that a,b are neither odd nor even so we again have an impossibility and we reject our assumption.

QED