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
2^5m5⁵ⁿx'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 5^5n-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 + 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:
2^5m5⁵ⁿx'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