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)=1and
55nx'5 = y5 + z5If 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
oddp = odd + even = oddq = odd - even = odd(c)
gcd(p,q)=1Assume 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
rWe 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
evena' = (odd)2 + 25(odd)2 = odd + odd = evenb' 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)=1a = (1/2)(q2 + 25r2)b = 5r2Assume there exists a prime
f that divides both
a,bThere are three cases that we need to consider:
Case I:
f = 2Case II:
f = 5Case III:
f divides both
q,rCase 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
bFrom 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 + 25r2 ≡
1 + (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 = 5Case 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)/16b = 5d(c4 + 10c2d2 + 5d4)/16gcd(c,d)=1c,d are both odd
5 doesn't divide
c
5 divides
dc,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') = 1c',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