## Monday, October 31, 2005

### Fermat's Last Theorem: n=5 : 5 doesn't divide z

For those interested in the history behind this proof, you may want to start here.

For those who would like to start at the beginning of this proof, start here.

The proof presented is based on two books: Harold M. Edward's Fermat's Last Theorem: A Genetic Introduction and Paulo Ribenboim's Fermat's Last Theorem for Amateurs.

Lemma: There are no integers x,y,z such that x5 + y5 = z5, xyz ≠ 0, gcd(x,y,z)=1, x,y are odd, z is even, and 5 divides x or y.

(1) Assume that x,y,z exist.

(2) For purposes of this proof, we will assume that 5 divides x (if it divides y, then we can make the exact same arguments replacing y for x)

(3) We know that there exists n,x' such that [since x is divisible by 5 from (#2)]:
x = 5nx' with n ≥ 1, gcd(x',5)=1
and 55nx'5 = y5 + z5

If we multiply 25 to both sides, we get:
(2)5(5)5nx'5 = 25(y5 + z5)

(4) from this we know that there are two integers p,q that have the following properties:
gcd(p,q)=1
p,q are both odd
(2)5(5)5nx'5 = 2p(p4 + 10p2q2 + 5q4)
p,q ≠ 0
(a) Let p = y + z, q = y - z

(b) p,q are both odd

p = odd + even = odd
q = odd - even = odd

(c) gcd(p,q)=1

Assume there is a prime f that divides both p,q such that p = fp', q = fq'. f is odd since p,q are odd.

f would divide y since p + q = f(p' + q') = y + z + y - z = 2y and since f is odd.

But f would also divide z since p - q = f(p' - q') = y + z - y + z = 2z and since f is odd.

But this is impossible since gcd(y,z)=1.

(d) (2)5(5)5nx'5 = 2p(p4 + 10p2q2 + 5q4) since:

(2)5(5)5nx'5 = 25(y5 + z5) =
(2y)5 + (2z)5 = (p+q)5 + (p-q)5 =
= 2p(p4 + 10p2q2 + 5q4) [By Lemma 1, here]

(e) p,q ≠ 0 since gcd(y,z)=1. [If y+z=0 or (y-z)=0, then y=z or y=-z which is not possible.]

(5) There exists an integer r that has these properties:
p=5r
gcd(q,r)=1
q,r are both odd
25m55nx'5 = 2*52r(q4 + 50q2 r2 + 125r4)
5 divides r
r ≠ 0
(a) 5 divides p since:

From step (#4d) and Euclid's Generalized Lemma, 5 divides 2p or 5 divides p4 + 10p2q2 + 5q4. If it divides 2p, then it divides p. Likewise, if 5 divides p4 + 10p2q2 + 5q4 then it must also divide p. [Details if needed are here] Either way, it divides p.

(b) Therefore, there exists a value r such that:
p = 5r

(c) gcd(r,q)=1 since gcd(p,q)=1 from step (#4c).

(d) We know r,q are both odd since:

odd divided by 5 = odd

(e) Finally, we note that applying r to step #4e gives us:
2p(p4 + 10p2q2 + 5q4)=
2(5r)[(5r)4 + 10(5r)2q2 + 5q4] =
2*5r*5(125r4 + 50r2q2 + q4) =
2*52r(q4 + 50q2 r2 + 125r4)

(f) 5 divides r

We know that 55n divides 2 * 52r(q4 + 50q2r2 + 125r4) from step (#4e and #5e).

So that 55n-2 divides 2 * r(q4 + 50q2r2 + 125r4)

Since n ≥ 1 (#2), 5n ≥ 5 which means 5n is greater than 2 so that we know that:
5 divides 2 * r(q4 + 50q2r2 + 125r4)

Now, we know that 5 doesn't divide q (since 5 divides p and gcd(p,q)=1 ) so that 5 cannot divide q4 + 50q2r2 + 125r4.

By Euclid's Generalized Lemma, since 5 divides 2 * r, it must divide r.

(g) r ≠ 0 since p ≠ 0 from step #4f.

(6) We know that there exists three integers a,b,t since:

(a) Let t' = q4 + 50q2 r2 + 125r4

(b) Let a' = q2 + 25r2

(c) Let b' = 10r2

(d) And we note that t' = a'2 - 5b'2. [By Lemma 2, here]

(e) Now a', b' are even

a' = (odd)2 + 25(odd)2 = odd + odd = even
b' is a multiple of 10.

(f) So, we know that there exists a,b such that:
a = (1/2)a'
b = (1/2)b'

(g) Let t = (1/4)t' = (1/4)(a'2 - 5b'2) = [(1/2)a']2 - 5[(1/2)b']2 = a - 5b2

(7) a,b,t have the following properties:

(a) gcd(a,b)=1

a = (1/2)(q2 + 25r2)
b = 5r2

Assume there exists a prime f that divides both a,b

There are three cases that we need to consider:
Case I: f = 2
Case II: f = 5
Case III: f divides both q,r

Case I isn't true since b is odd since r is odd.

Case II isn't true since 5 doesn't divide a. Multiplying 2 to both sides and using Euclid's Lemma, we know that if 5 divides a, then it would need to divide q2 + 25r2. But 5 doesn't divide q since 5 divides p and gcd(p,q)=1 so 5 cannot divide q2 + 25r2.

Case III isn't true since gcd(q,r)=1 from Step #4c.

So, they cannot have any common divisors.

(b) 5 doesn't divide a, 5 divides b

From step #6a we know that 5 divides b. From (a), we know that gcd(a,b)=1 so 5 can't divide a.

(c) a,b are both odd.

We already showed that b is odd so all we need to do is show that a is odd.

If a value is odd, there exists a value u such that: odd = 2u+1
(odd)2 = (2u + 1)2 = 4u2 + 4u + 1
So any (odd)2 ≡ 1 (mod 4) [start here if you need a review of modular arithmetic]
a'q2 + 25r21 + (1)(1) ≡ 2 (mod 4)

So, if a' ≡ 2 (mod 4), that there exists a value v such that a' - 2 = 4v and that (2a) -2 = 4v which means a - 1 = 2v and a = 2v + 1.

(d) (52r) * (t/4) = a fifth power since:

From #4e, we have:
25m55nx'5 = 2*52r(q4 + 50q2 r2 + 125r4)

Applying #5, we get
(2)5(5)5nx'5 = 2*52rt' = 2352rt

And dividing both sides by 25 gives us:
55nx'5 = (52rt)/4

(e) gcd(52r ,t/4) = 1 since:

Assume that there is a prime f that divides both. There are two cases that we need to consider:
Case I: f = 5
Case II: f divides r and t and f is odd (since r is odd)

Case I is false since 5 doesn't divide t'. t' = q4 + 50q2r2 + 125r4 but 5 can't divide t' since 5 doesn't divide q (see #6a).

Case II is false since f would need to divide q in order to divide t but gcd(r,q)=1 (#4c).

(f) t/4=(1/4)(a - 5b2 ) is a fifth power since:

#6d, #6e and Relatively Prime Divisors of n-powers.

(g) a,b are positive integers

We know this since q,r are nonzero integers (#3f,#4g) and #5 (since the square of a nonzero integer is a positive integer).

(8) Applying a lemma (see Lemma 2, here) , we can conclude that there exist c,d such that:

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

(9) 53a =(1/4) (54d)[([c2 + 5d2]/2)2 - 5d4] since:

53a = (53)(5d)(c4 + 10c2d2 + 5d4)/16 = [(54d)/4][(c4 + 10c2d2 + 5d4)/4]

[(c2 + 5d2)/2]2 = (c4 + 10c2d2 + 25d4)/4

[(c2 + 5d2)/2]2 - 5d4 =
(c4 + 10c2d2 + 25d4)/4 - 20d4/4 = c4 + 10c2d2 + 5d4)/4

(10) [(c2 + 5d2)/2]2 - 5d4 ≡ 0 (mod 4)

(a) c2 ≡ 1 (mod 4) [since c is odd (#8) and step (#7c)]

(b) 5d2 ≡ (1)(1) ≡ 1 (mod 4) [since d is odd (#8) and step (#7c)]

(c) 5d4 = 5(d2)2 ≡ (1)(1) ≡ 1 (mod 4) [see above]

(d) [(c2 + 5d2)/2]2 - 5d4 ≡ [(1 + 1)/2][(1 + 1)/2] - 1 ≡ (1)(1) - 1 ≡ 0 (mod 4) [Note: for a review of modular arithmetic, go here]

(11) gcd([54d]/4 , [(c2 + 5d2)/2]2 - 5d4) = 1 since:

(a) Assume there is a prime f that divides both (54d)/4 and [(c2 + 5d2)/2]2 - 5d4.

(b) By Euclid's Lemma, f is either 5 or it divides d.

(c) f ≠ 5 since 5 doesn't divide c (#8) and therefore cannot divide [(c2 + 5d2)/2]2 - 5d4. [See here]

(d) f cannot divide d because if it did, then it would divide both d and [(c2 + 5d2)/2]2 - 5d4 which means that it would also divide c but this is impossible since gcd(c,d)=1 (#8).

(12) So, we know that (54d and (1/4)([(c2 + 5d2)/2]2 - 5d4) are fifth powers by Relatively Prime Divisors of n-powers theorem.

(13) Applying a lemma (see Lemma 2, here), we know that there exists c', d' such that:

(c2 + 5d2)/2 = c'(c'4 + 50c'2d'2 + 125d'4)

d2 = 5d'(c'4 + 10c'2d'2 + 5d'4)/16.

gcd(c',d') = 1

c',d' are both odd

5 doesn't divide c'

5
divides d'

c', d' ≠ 0

(14) We note that (58)d2 = (1/4)(59d')[([c'2 + 5d'2]/2)2 - 5(d'2)2] since:

(58)d2 = (59)d'(c'4 + 10c'2d'2 + 5d'4)/16 =
[(59)d'/4][(c'4 + 10c'2d'2 + 5d'4)/4]

(c'2 + 5d'2)2 = c'4 + 10c'2d'2 + 25d'2

(c'4 + 10c'2d'2 + 5d'4)/4 = [(c'2 + 5d'2)2 - 20d'2]/4 = [(c'2 + 5d'2)/2]2 - 5(d'2)2

(15) gcd((59)d' , (1/4)[(c'2 + 5d'2)/2]2 - 5(d'2)2 ) = 1 since:

(a) We can divide it up in this way since d' is odd, we know that 4 must divide the other value.

(b) Assume there is a prime f that divides both

(c) f = 5 or f divides d'

(c) f ≠ 5 since 5 doesn't divide c' (#13)

(d) f doesn't divide d' because if it did and it divided (1/4)[(c'2 + 5d'2)/2]2 - 5(d'2)2 then it would divide c' but gcd(c',d') = 1 (#13)

(16) We note that 58d2 is a fifth power since:

58d2 = (54d)2 and 54d is a fifth power (#12) [Since (x5)2 = (x2)5]

(17) So, we can conclude that 59d' and (1/4)[(c'2 + 5d'2)/2]2 - 5(d'2)2 are fifth powers by the Relatively Prime Divisors of N-Powers Theorem.

(18) Finally, we note that d' is ≥ 1 and less than d since:

(a) 25d'5 is less than 16d2 from step (#13)

(b) d' is greater than 0 (#13)

(c) Assume that d' ≥ d

(d) In this case, 25d'5 would be greater than 16d2 since d',d are integers, 25 is greater than 16 and d'5 is greater than d'2 which is greater or equal to d2.

(e) But this is not the case so we can reject our assumption at (c).

(19) But this means that we can repeat this process d times using the lemma stated above (see Lemma 2, here) and eventually get an integer d'' that is less than 1 and greater than 0 which is impossible.

QED

Scouse Rob said...

In step (9) should:

5^3a

be

5^3b

Rob

Scouse Rob said...

It took a bit of work to figure out step (12).

Is it because of the following?

5^2r is a power of 5 from step (7d, 7e)

b=5r^2

5^3b = 5^4r^2 = (5^2r)^2

So 5^3b is a 5th power.

Rob