Fermat's Last Theorem: n=5: 5 divides 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 z.

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

(2) We know that there exists m,n,z' such that:
z = 2^m5ⁿz' with m ≥ 1, n ≥ 1, [Since 5 divides z and z is even]
gcd(z',2)=1, gcd(z',5)=1. [Since we can make m and n high enough to cover all 2's and 5's]
and 2^5m5⁵ⁿz'⁵ = x⁵ + y⁵ [Since z=2^m5ⁿz', z⁵ = 2^5m5⁵ⁿz'⁵]

(3) There exist two integers p,q that have the following properties:

gcd(p,q)=1
p,q have different parities (one is odd, one is even)
2^5m5⁵ⁿz'⁵ = 2p(p⁴ + 10p²q² + 5q⁴)
p,q ≠ 0

since:

(a) x,y are odd → x+y is even, x - y is even

Since odd + odd = even and odd - odd = even.

[We know that x,y are odd since gcd(x,y,z)=1 and z is even]

(b) From this, we know that there are two values p,q such that:
x + y = 2p
x - y = 2q

(c) We also know that gcd(p,q)=1 since:

x = (1/2)[x + y + x - y] = (1/2)(2x) = (1/2)[2p + 2q] = p + q
y = (1/2)[x + y - (x - y)] = (1/2)(2y) = (1/2)[2p - 2q] = p - q

If there exists a value d greater than 1 that divides both p,q then, from above, it would divide both x,y which is impossible since gcd(x,y)=1 from step (#1).

(d) From step 3(c) above, we can conclude that p,q have different parities (that is, one is odd and the other is even).

p,q can't both have even parity because then x would be even by step 3(c) above since even + even = even but x is odd by step 3(a) above.

p,q can't both have odd parity for the same reason since odd + odd = even.

Therefore, p,q must have different parities where one is odd and one is even since odd + even = odd.

(e) Applying p,q to step (#2) above gives us:

2^5m5⁵ⁿz'⁵ = x⁵ + y⁵ = (p+q)⁵ + (p-q)⁵ = 2p(p⁴ + 10p²q² + 5q⁴) [See Lemma 1, here]

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

(4) There exists an integer r that has these properties:

p=5r
gcd(q,r)=1
q,r have different parities (one is odd, one is even)
2^5m5⁵ⁿz'⁵ = 2*5²r(q⁴ + 50q² r² + 125r⁴)
5 divides r
r ≠ 0

since:

(a) 5 divides p since:

From step (#3e) 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) From (a), we know that there must exist r such that:
p = 5r

(c) gcd(r,q)=1 since gcd(p,q)=1 from step (#3c).

(d) We know r,q have different parities since r has the same parity as p since:

odd divided by 5 = odd
even divided by 5 = even

p,q have different parities (from step #3d)

(e) Finally, we note that applying r to step #3e 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 since:

We know that 5⁵ⁿ divides 2 * 5²r(q⁴ + 50q²r² + 125r⁴) from step (#3e and #4e).

So that 5^5n-2 divides 2 * r(q⁴ + 50q²r² + 125r⁴)

Since n ≥ 1 (step #2 above), 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 #3f.

(5) We define three values a,b,t to be the following:

(a) Let t = q⁴ + 50q² r² + 125r⁴

(b) Let a = q² + 25r²

(c) Let b = 10r²

(d) And we note that t = a² - 5b². [By Lemma 2, here]

(6) Now, we note that a,b have the following properties:

(a) gcd(a,b)=1 since:

Assume gcd(a,b) is greater than 1.

Then, there is a prime f that divides both a and b.

Since f divides a, we know that f divides 10r² which by Euclid's Generalized Lemma, gives us three possible cases:

Case I: f divides 2 (in this case, f = 2)
Case II: f divides 5 (in this case, f = 5)
Case III: f divides r (since f divides r² implies f divides r or f divides r.)

Case I is false: because a is odd. a must be odd because q,r have different parities (step #4d) and since (odd)² + 25(even)² = odd + even = odd and (even)² + 25(odd)² = even + odd = odd.

Case II is false: f can't be 5 since 5 divides p and gcd(p,q)=1. This means 5 doesn't divide q and therefore 5 doesn't divide q² + 25r².

Case III is false: If f divides r and f divides a, then f would divide q (since a= q² + 25r² and see here) but this cannot be the case since gcd(r,q)=1.

Since all cases are false, no such prime can exist.

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

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

(c) a,b have different parities

b is even from #5c so that 2 divides b. But 2 cannot divide a since gcd(a,b)=1.

(d) gcd(2*5²r,t) = 1 since:

Assume there exists a prime f that divides both 2*5²r and t.

We know that f ≠ 2 since t is odd since:

t = (q² + 25r²)² - 5(10r²)².
q,r have different parity (#4d)

Case I: q is odd, r is even
(odd² + 25(even)²)² - 5(10(even)²)² = (odd + 25(even))² - 5(even)² =
= (odd + even)² - even = odd² - even = odd
So, if q is odd, r is even, then t is odd.

Case II: q is even, r is odd
(even² + 25(odd)²)² - 5(10(odd)²)² = (even + odd)² - 5(even)² =
odd² - even = odd.
So, if q is even, r is odd, then t is odd.

Either way, t is odd.

We know that f ≠ 5 since t is not divisible by 5 since 5 doesn't divide q (#6a) [Additional details if needed are here]

Finally, f cannot divide r since gcd(r,t)=1 since:

Assume there exists a prime p' that divides both t,r
p' would then divide q [Additional details if needed are here]
But this is impossible since gcd(q,r)=1 from (#4c).

Therefore, the prime p' doesn't exist and we can conclude that f does not divide r.

(e) t=a - 5b² is a fifth power since:

Since z⁵ = (2*5²r)*(t) from step (#4e) above and since (2*5²r) and (t) are relatively prime from step (# 6d) above, we can conclude that:
t is a fifth power and (2*5²r) is a fifth power [From Relatively Prime Divisors of n-powers]

(f) 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).

(7) With the properties in step #6, we can use a lemma (see Lemma 1, here):

There exists two integers c,d such that:

a = c(c⁴ + 50c²d² + 125d⁴)
b = 5d(c⁴ + 10c²d² + 5d⁴)
gcd(c,d)=1
c,d have different parities
5 doesn't divide c
5 divides d
c,d are nonzero

(8) Let u' = c + 5d², v' = 2d²

(9) We note that u', v' have the following properties:

(a) gcd(u',v')=1

Assume there is a prime f that divides both c + 5d², 2d².

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

Case I isn't true because u' = c + 5d² is odd since c,d have different parities (#8) and in both cases, u' is odd:

(odd) + 5(even)² = odd + even = odd
(even) + 5(odd)² = even + odd = odd

Case II isn't true 5 doesn't divide c (#8)

Finally Case III isn't true since gcd(c,d)=1 (#8)

(b) 5 doesn't divide u', 5 divides v'

We know that 5 divides d (#8b) so 5 divides v'=2d². 5 doesn't divide u' since gcd(u',v')=1 (#8)

(c) u',v' have different parities

v' is even by definition. We showed already that u' is odd (#10a)

(d) u' - 5v'² is a fifth power since:

c⁴ + 10c²d² + 5d⁴ can be put into the form (c² + 5d²)² - 5(2d²)² since:

(c² + 5d²)² = c⁴ + 10c²d² + 25d⁴ [From the Binomial Theorem]

5(2d²)² = -20d⁴

Now, (2*5²r) is a fifth power (#7e) so (2*5²r)² is a fifth power [since (x⁵)² = (x²)⁵]

(2*5²r)² = 2*5³*10r² = (2*5³)*v = (2*5³)[5d(c⁴ + 10c²d² + 5d⁴)] =
(2*5⁴d)(c⁴ + 10c²d² + 5d⁴)

So, we next show that gcd( 2*5⁴d ,c⁴ + 10c²d² + 5d⁴ ) = 1

Assume that there is a prime f that divides both. We need handle three cases:

Case I: f = 2
Case II: f=5
Case III: f divides both c,d

Case I is impossible since c⁴ + 10c²d² + 5d⁴ is odd since c,d have different parities (#8) and:
(odd)⁴ + 10(odd)²(even)² + 5(even)⁴ = odd + even + even = odd
(even)⁴ + 10(even)²(odd)² + 5(odd)⁴ = even + even + odd = odd

Case II is impossible since 5 doesn't divide c (#8)

Case III is impossible since gcd(c,d)=1 (#8)

OK, combining these results with Relatively Prime Divisors of n-powers, we can conclude that:
2*5⁴d and c⁴ + 10c²d² + 5d⁴ are 5th powers.

(10) With the properties in step #9, we can use a lemma which I will prove later to show:

c + 5d² = c'(c'⁴ + 50c'²d'² + 125d'⁴)

2d² = 5d'(c'⁴ + 10c'²d'² + 5d'⁴)

gcd(c',d')=1

c',d' have different parities

5 doesn't divide c'

5 divides d'

c' , d' ≠ 0

(11) (2*5⁸)(2d²) = 2²*5⁸d² = (2*5⁴d)²

(12) We know that 2*5⁴d is a fifth power from step (#9d) so (2*5⁴d)² is also a fifth power [Because (x⁵)² = (x²)⁵]

(13) So 2*5⁹d'(c'⁴ + 10c'²d'² + 5d'⁴) is a fifth power by the argument below:

First, we note that (2*5⁸)(2d²) is a fifth power since:
(2*5⁸)(2d²) = (2*5⁴d)² (from step #11) and (2*5⁴d)² is a fifth power (from step #12)

But this shows that 2*5⁹d'(c'⁴ + 10c'²d'² + 5d'⁴) is a fifth power since:
2*5⁹d'(c'⁴ + 10c'²d'² + 5d'⁴) = (2*5⁸) * 5d'(c'⁴ + 10c'²d'² + 5d'⁴) = (2*5⁸)(2d²) [from step #10.]

(14) gcd(2*5⁹d' , c'⁴ + 10c'²d'² + 5d'⁴)=1 since:

(a) Assume there is a prime f that divides both 2*5⁹d' and c'⁴ + 10c'²d'² + 5d'⁴.

(b) There are three cases that we need to consider by Euclid's Generalized Lemma:

Case I: f = 2
Case II: f = 5
Case III: f divides d' and c'⁴ + 10c'²d'² + 5d'⁴

(c) We can eliminate Case I since is c'⁴ + 10c'²d'² + 5d'⁴ odd.

c',d' have opposite parity (#10) which means that the value is odd

(even)⁴ + 10(even)²(odd)² + 5(odd)⁴ = even + even + odd = odd

(odd)⁴ + 10(odd)²(even)² + 5(even)⁴ = odd + even + even = odd

(d) We can eliminate Case II since 5 doesn't divide c' which 5 cannot divide c'⁴ + 10c'²d'² + 5d'4. [Detail if needed is here]

(e) We can eliminate Case III since gcd(c',d') = 1. [From step (#10)]

(15) So 2*5⁹d' and c'⁴ + 10c'²d'² + 5d'⁴ are also fifth powers. [By (#14) and Relatively Prime Divisors of n-powers]

(16) Step (#15) puts us in the exact same position as Step (#9). This means that using the same argument, we can apply the lemma (see Lemma 1, here) as many times as we would like.

(17) Moreover, d' is greater than 0 and d' is less than d since:

25d'⁵ ≤ 5d'(c'⁴ + 10c'²d'² + 5d'⁴) = 2d² [From steps (#10) since c',d' are nonzero and therefore c'² and d'² ≥ 1]

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

Also:

25d'⁵ ≤ 2d² → d'⁵ ≤ (2d²)/25 → d' ≤ ⁵√(2d*2d)/25

So:

d' is less than d [Since ⁵√(2d*2d)/25 is less than d as shown below]

⁵√(2d*2d)/25 is less than d since

(i) Putting both sides to the power of 5 gives us:
(2d²)/25 is less than d⁵

(ii) Multiplying both sides by 25 gives us:
2d² is less than 25d⁵

(iii) Since d ≥ 1 (#7), we know that d² is less than d⁵.

(18) If this procedure continued, we eventually get an integer d'' which is less than 1 which is absurd [We only need to repeat this procedure d more times]

QED