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 x

^{5}+ y

^{5}= z

^{5}, 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 = 5

^{n}x' with n ≥ 1, gcd(x',5)=1

and 5

^{5n}x'

^{5}= y

^{5}+ z

^{5}

If we multiply 2

^{5}to both sides, we get:

(2)

^{5}(5)

^{5n}x'

^{5}= 2

^{5}(y

^{5}+ z

^{5})

(4) from this we know that there are two integers p,q that have the following properties:

gcd(p,q)=1(a) Let p = y + z, q = y - z

p,q are both odd

(2)^{5}(5)^{5n}x'^{5}= 2p(p^{4}+ 10p^{2}q^{2}+ 5q^{4})

p,q ≠ 0

(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)

^{5n}x'

^{5}= 2p(p

^{4}+ 10p

^{2}q

^{2}+ 5q

^{4}) since:

(2)

^{5}(5)

^{5n}x'

^{5}= 2

^{5}(y

^{5}+ z

^{5}) =

(2y)

^{5}+ (2z)

^{5}= (p+q)

^{5}+ (p-q)

^{5}=

= 2p(p

^{4}+ 10p

^{2}q

^{2}+ 5q

^{4}) [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(a) 5 divides p since:

gcd(q,r)=1

q,r are both odd

2^{5m}5^{5n}x'^{5}= 2*5^{2}r(q^{4}+ 50q^{2}r^{2}+ 125r^{4})

5 divides r

r ≠ 0

From step (#4d) and Euclid's Generalized Lemma, 5 divides 2p or 5 divides p

^{4}+ 10p

^{2}q

^{2}+ 5q

^{4}. If it divides 2p, then it divides p. Likewise, if 5 divides p

^{4}+ 10p

^{2}q

^{2}+ 5q

^{4}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(p

^{4}+ 10p

^{2}q

^{2}+ 5q

^{4})=

2(5r)[(5r)

^{4}+ 10(5r)

^{2}q

^{2}+ 5q

^{4}] =

2*5r*5(125r

^{4}+ 50r

^{2}q

^{2}+ q

^{4}) =

2*5

^{2}r(q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4})

(f) 5 divides r

We know that 5

^{5n}divides 2 * 5

^{2}r(q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4}) from step (#4e and #5e).

So that 5

^{5n-2}divides 2 * r(q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4})

Since n ≥ 1 (#2), 5n ≥ 5 which means 5n is greater than 2 so that we know that:

5 divides 2 * r(q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4})

Now, we know that 5 doesn't divide q (since 5 divides p and gcd(p,q)=1 ) so that 5 cannot divide q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4}.

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' = q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4}

(b) Let a' = q

^{2}+ 25r

^{2}

(c) Let b' = 10r

^{2}

(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 - 5b

^{2}

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

(a) gcd(a,b)=1

a = (1/2)(q

^{2}+ 25r

^{2})

b = 5r

^{2}

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 q

^{2}+ 25r

^{2}. But 5 doesn't divide q since 5 divides p and gcd(p,q)=1 so 5 cannot divide q

^{2}+ 25r

^{2}.

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}= 4u

^{2}+ 4u + 1

So any (odd)

^{2}≡ 1 (mod 4) [start here if you need a review of modular arithmetic]

a' ≡ q

^{2}+ 25r

^{2}≡ 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) (5

^{2}r) * (t/4) = a fifth power since:

From #4e, we have:

2

^{5m}5

^{5n}x'

^{5}= 2*5

^{2}r(q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4})

Applying #5, we get

(2)

^{5}(5)

^{5n}x'

^{5}= 2*5

^{2}rt' = 2

^{3}5

^{2}rt

And dividing both sides by 2

^{5}gives us:

5

^{5n}x'

^{5}= (5

^{2}rt)/4

(e) gcd(5

^{2}r ,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' = q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4}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 - 5b

^{2}) 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(c

^{4}+ 50c

^{2}d

^{2}+ 125d

^{4})/16

b = 5d(c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})/16

gcd(c,d)=1

c,d are both odd

5 doesn't divide c

5 divides d

c,d are nonzero.

(9) 5

^{3}a =(1/4) (5

^{4}d)[([c

^{2}+ 5d

^{2}]/2)

^{2}- 5d

^{4}] since:

5

^{3}a = (5

^{3})(5d)(c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})/16 = [(5

^{4}d)/4][(c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})/4]

[(c

^{2}+ 5d

^{2})/2]

^{2}= (c

^{4}+ 10c

^{2}d

^{2}+ 25d

^{4})/4

[(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}=

(c

^{4}+ 10c

^{2}d

^{2}+ 25d

^{4})/4 - 20d

^{4}/4 = c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})/4

(10) [(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}≡ 0 (mod 4)

(a) c

^{2}≡ 1 (mod 4) [since c is odd (#8) and step (#7c)]

(b) 5d

^{2}≡ (1)(1) ≡ 1 (mod 4) [since d is odd (#8) and step (#7c)]

(c) 5d

^{4}= 5(d

^{2})

^{2}≡ (1)(1) ≡ 1 (mod 4) [see above]

(d) [(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}≡ [(1 + 1)/2][(1 + 1)/2] - 1 ≡ (1)(1) - 1 ≡ 0 (mod 4) [Note: for a review of modular arithmetic, go here]

(11) gcd([5

^{4}d]/4 , [(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}) = 1 since:

(a) Assume there is a prime f that divides both (5

^{4}d)/4 and [(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}.

(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 [(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}. [See here]

(d) f cannot divide d because if it did, then it would divide both d and [(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}which means that it would also divide c but this is impossible since gcd(c,d)=1 (#8).

(12) So, we know that (5

^{4}d and (1/4)([(c

^{2}+ 5d

^{2})/2]

^{2}- 5d

^{4}) 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:

(c

^{2}+ 5d

^{2})/2 = c'(c'

^{4}+ 50c'

^{2}d'

^{2}+ 125d'

^{4})

d

^{2}= 5d'(c'

^{4}+ 10c'

^{2}d'

^{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 (5

^{8})d

^{2}= (1/4)(5

^{9}d')[([c'

^{2}+ 5d'

^{2}]/2)

^{2}- 5(d'

^{2})

^{2}] since:

(5

^{8})d

^{2}= (5

^{9})d'(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4})/16 =

[(5

^{9})d'/4][(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4})/4]

(c'

^{2}+ 5d'

^{2})

^{2}= c'

^{4}+ 10c'

^{2}d'

^{2}+ 25d'

^{2}

(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4})/4 = [(c'

^{2}+ 5d'

^{2})

^{2}- 20d'

^{2}]/4 = [(c'

^{2}+ 5d'

^{2})/2]

^{2}- 5(d'

^{2})

^{2}

(15) gcd((5

^{9})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 5

^{8}d

^{2}is a fifth power since:

5

^{8}d

^{2}= (5

^{4}d)

^{2}and 5

^{4}d is a fifth power (#12) [Since (x

^{5})

^{2}= (x

^{2})

^{5}]

(17) So, we can conclude that 5

^{9}d' 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 16d

^{2}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 16d

^{2}since d',d are integers, 25 is greater than 16 and d'

^{5}is greater than d'

^{2}which is greater or equal to d

^{2}.

(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

## 2 comments:

In step (9) should:

5^3abe

5^3bRob

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

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