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 = a^{2} + 1 and a ≥ 1, then the continued fraction expansion of √d is [a , 2a, ... ](1)

a_{0} = a since a is less than

√a^{2} + 1 which is less than

a + 1 since:

a =√a^{2}a + 1 = √a^{2} + 2a + 1a^{2} is less than

a^{2} + 1 is less than

a^{2} + 2a + 1.

(2)

a_{2} = 2a since

2a is less than

1/(√a^{2} + 1 - a) which is less than

2a + 1.

(a)

1/(√a^{2} + 1 - a) = (√a^{2} + 1 + a)/(a^{2} + 1 - a^{2}) = √a^{2} + 1 + a.

(b) From (a), we see that

2a is less than

a + √a^{2} + 1 since

a^{2} + 1 is greater than

a^{2}.

(c) Since

a^{2} + 1 is less than

a^{2} + 2a + 1, we see that:

√a^{2} + 1 + a is less than

a + (a + 1) = a + √a^{2} + 2a + 1(d) Finally, we see that all other items from this point

a_{i} are =

2a.

1/(√a^{2} + 1 + a - 2a) = 1/(√a^{2} + 1 - a) == (√a^{2} + 1 + a)/(a^{2} + 1 - a^{2}) = √a^{2} + 1 + a.

QED

Lemma 2: if p_{k}/q_{k} is a convergent for the continued fraction expansion of Î±, 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_{k}q_{k-1} - p_{k-1}q_{k} = (-1)^{k-1} (see Lemma 2

here)

(3) From (1), we know that

f divides

p_{k}q_{k-1} - p_{k-1}q_{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) Î± is an irrational number(b) r,s are integers(c) s is greater than 0(d) absolute(sÎ± - r) is less than absolute(q_{k}Î± - p_{k}) where p_{k},q_{k} are the convergents of the continued fraction expansion of Î±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_{k}x + p_{k+1}y = rq_{k}x + q_{k+1}y = s(3) Multiplying

q_{k} to the first equation and

p_{k} to the second gives us:

p_{k}q_{k}x + p_{k+1}q_{k}y = rq_{k}p_{k}q_{k}x + p_{k}q_{k+1}y = sp_{k}(4) Now, subtracting the second from the first gives us:

y(p_{k+1}qk - p_{k}qk+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_{k}q_{k+1} - p_{k+1}q_{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_{k}x(c) Then

s = q_{k}x(d) So that

absolute(sÎ± - r) = absolute(x) * absolute(q_{k}Î± - p_{k})(e) Because

x ≠ 0, we then have

absolute(x) * absolute(q_{k}Î±-p_{k}) ≥ absolute(q_{k}Î± - p_{k})(f) But combining (d) with (e) gives us:

absolute(sÎ± - r) ≥ absolute(q_{k}Î± - 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_{k}x = s - q_{k+1}y. 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+1}y ≥ q_{k+1} which is greater than

s (by assumption #1) so

q_{k}x = s - q_{k+1}y is less than

0.

(10) We know that

Î± 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

Î± which is less

(p_{k+1})/q_{k+1}Which means that

p_{k} is less than

Î±q_{k} and that

Î±q_{k+1} is less than

p_{k+1}.

This then gives us that:

q_{k}Î± - p_{k} is positive and

q_{k+1}Î± - p_{k+1} is negative.

(b) We also know that if

k is odd,

(p_{k+1})/q_{k+1} is less than

Î± which is less than

p_{k}/q_{k}Which means that

p_{k+1} is less than

Î±q_{k+1} and

q_{k+1}Î± is less than

p_{k}.

This then gives us that:

q_{k+1}Î± - p_{k+1} is positive and

q_{k}Î± - p_{k} is negative.

(11) By 10(a) and 10(b), we know that regardless of whether

k is odd or even,

q_{k}Î± - p_{k} and

q_{k+1}Î± - p_{k+1} have opposite signs.

(12) This means that

x(q_{k}Î± - p_{k}) and

y(q_{k+1}Î± - p_{k+1}) have the same sign.

(13) Finally, this gives us that:

absolute(sÎ± - r) == absolute( [q_{k}x + q_{k+1}y]Î± - [p_{k}x + p_{k+1}y]) == absolute( x[q_{k}Î± - p_{k}] + y[q_{k+1}Î± - p_{k+1}] )(14) This then gives us:

absolute(sÎ± - r) ≥ absolute(x)*absolute(q_{k}Î± - p_{k}) + absolute(y)*absolute(q_{k+1}Î± - p_{k+1}) ==>absolute(sÎ± - r) ≥ absolute(x)*absolute(q_{k}Î± - p_{k}) ==>absolute(sÎ± - r) ≥ absolute(q_{k}Î± - 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_{k}Î± - p_{k}) ≤ absolute(sÎ± - r).(1) Assume that

absolute(sÎ± - r) is less than

absolute(q_{k}Î± - 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(Î± - r/s) is less than 1/(2s^{2}) and s is greater than 0, then r/s is a convergent of the continued fraction of Î±(1) Assume that

r/s is not a convergent of

Î±(2) Then,

r/s ≠ p_{n}/q_{n} for all

n.

(3) Since

q_{0} = 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(Î± - r/s) is less than

1/(2s^{2}) implies that:

absolute(sÎ± - r) is less than

1/2s.

(6) From a result in Lemma 4 above,

absolute(q_{k}Î± - p_{k}) ≤ absolute(sÎ± - r).

Which then gives us:

absolute(q_{k} Î± - p_{k}) is less than

1/2s.

(7) Dividing all sides by

q_{k} gives:

absolute(Î± - 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 + Î± - Î± )which is:

≤ absolute(Î± - p_{k}/q_{k}) + absolute(Î± - r/s)which is less than:

1/(2sq_{k}) + 1/2s^{2}(10) Putting it all together gives us:

1/sq_{k} is less than

1/(2sq_{k}) + 1/(2s^{2})(11) Subtracting

1/2q_{k} from both sides gives us:

1/(2sq_{k}) is less than

1/(2s^{2}).(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 x^{2} - dy^{2} 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:

x^{2} - dy^{2} = (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 == (x^{2} - dy^{2})/[y(x + y√d)](5) This is less than:

(x^{2} - dy^{2})/[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

(x^{2} - dy^{2})/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/(2y^{2})(8) Which by Lemma 5 above gives us our conclusion.

QED

Lemma 7: if x^{2} - dy^{2} is a negative number greater than -√d, then x,y are convergents in the continued fraction expansion of √d.(1)

y^{2} - (1/d)x^{2} is a positive number less than

1/√d since:

(a)

√d is greater than

dy^{2} - x^{2} (By multiplying by

-1 to the assumption)

(b)

√d/d is greater than

y^{2} - (x^{2})/d which is greater than

0.

(c)

1/√d is greater than

y^{2} -(x^{2})/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 == [y^{2} - x^{2}/d]/[x(y + x/√d)]which is less than:

[y^{2} - x^{2}/d]/(2x^{2}/√d) since:

(a)

y is greater than

x/√d (from #2 above) so

xy is greater than

x^{2}/√d.

(b) And

xy + x^{2}/√d is greater than

2x^{2}/√d.

(5) Now:

y^{2} - (x^{2}/d) is less than

1/√d (from #1) so:

(y^{2} - x^{2}/√d)/(2x^{2}/√d) is less than:

(1/√d)/(2x^{2}/√d)(6) Since:

(1/√d)/(2x^{2}/√d) = 1/2x^{2}, we have the following:

y/x - 1/√d is less than

1/2x^{2}.

(7) By the lemma above,

y/x is a convergent.

Lemma 8: if d = a^{2} + 1 and absolute(u^{2} - dv^{2}) ≠ 0,1, then absolute(u^{2} - dv^{2}) is greater than √d(1) The continued fraction for

√a^{2} + 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_{k}^{2} - dq^{2} = (-1)^{k}(4) Now if

absolute(u^{2} - dv^{2}) is less than

√d, then

u,v are convergents since:

(a) If

u^{2} - dv^{2} is less than

√d, then from Lemma 6 above,

u,v are convergents.

(b) If

-(u^{2} - dv^{2}) is less than

√d, then from Lemma 7 above,

u,v are convergents.

(5) From #4, since we are assuming that

absolute(u^{2} - dv^{2}) ≠ 1 or

0, then we can conclude that:

absolute(u^{2} - dv^{2}) 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] = (a^{2} - 5b^{2})/4(7) This means that:

(a^{2} - 5b^{2})/4 = ± 2 which implies that

a^{2} - 5b^{2} = ± 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 = 2ub = 2v(c)

(2u)^{2} - 5(2v)^{2} = ± 8So:

4u^{2} -20v^{2} = 4(u^{2} - 5v^{2}) = ± 8And:

u^{2} - 5v^{2} = ± 2.

(d) But

absolute(u^{2} - 5v^{2}) ≠ ± 2 from Lemma 8 above since

5 = 2^{2} + 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 + 1b = 2v + 1(c)

(2u+1)^{2} - 5(2v+1)^{2} = 4u^{2} + 4u + 1 - 5(4v^{2} + 4v + 1) == 4u^{2} + 4u + 1 - 20v^{2} -20v - 5 == 4u^{2} - 4u - 4 - 20v^{2} - 20v == 4(u^{2} + u - 1 - 5v^{2} - 5v) = ± 8(d) So we can conclude that:

u^{2} + u - 1 -5v^{2} - 5v = ± 2And:

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