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 z.

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

(2) We know that there exists m,n,z' such that:

z = 2

^{m}5

^{n}z' 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

^{5m}5

^{5n}z'

^{5}= x

^{5}+ y

^{5}[Since z=2

^{m}5

^{n}z', z

^{5}= 2

^{5m}5

^{5n}z'

^{5}]

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

gcd(p,q)=1since:

p,q have different parities (one is odd, one is even)

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

p,q ≠ 0

(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

^{5m}5

^{5n}z'

^{5}= x

^{5}+ y

^{5 }= (p+q)

^{5}+ (p-q)

^{5}= 2p(p

^{4}+ 10p

^{2}q

^{2}+ 5q

^{4}) [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=5rsince:

gcd(q,r)=1

q,r have different parities (one is odd, one is even)

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

5 divides r

r ≠ 0

(a) 5 divides p since:

From step (#3e) 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) 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(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 since:

We know that 5

^{5n}divides 2 * 5

^{2}r(q

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4}) from step (#3e and #4e).

So that 5

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

^{4}+ 50q

^{2}r

^{2}+ 125r

^{4})

Since n ≥ 1 (step #2 above), 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 #3f.

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

(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]

(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 10r

^{2}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 r

^{2}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 q

^{2}+ 25r

^{2}.

Case III is false: If f divides r and f divides a, then f would divide q (since a= q

^{2}+ 25r

^{2}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*5

^{2}r,t) = 1 since:

Assume there exists a prime f that divides both 2*5

^{2}r and t.

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

t = (q

^{2}+ 25r

^{2})

^{2}- 5(10r

^{2})

^{2}.

q,r have different parity (#4d)

Case I: q is odd, r is even

(odd

^{2}+ 25(even)

^{2})

^{2}- 5(10(even)

^{2})

^{2}= (odd + 25(even))

^{2}- 5(even)

^{2}=

= (odd + even)

^{2}- even = odd

^{2}- even = odd

So, if q is odd, r is even, then t is odd.

Case II: q is even, r is odd

(even

^{2}+ 25(odd)

^{2})

^{2}- 5(10(odd)

^{2})

^{2}= (even + odd)

^{2}- 5(even)

^{2}=

odd

^{2}- 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 - 5b

^{2}is a fifth power since:

Since z

^{5}= (2*5

^{2}r)*(t) from step (#4e) above and since (2*5

^{2}r) and (t) are relatively prime from step (# 6d) above, we can conclude that:

t is a fifth power and (2*5

^{2}r) 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(c

^{4}+ 50c

^{2}d

^{2}+ 125d

^{4})

b = 5d(c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})

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 + 5d

^{2}, v' = 2d

^{2}

(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 + 5d

^{2}, 2d

^{2}.

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 + 5d

^{2}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'=2d

^{2}. 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:

c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4}can be put into the form (c

^{2}+ 5d

^{2})

^{2}- 5(2d

^{2})

^{2}since:

(c

^{2}+ 5d

^{2})

^{2}= c

^{4}+ 10c

^{2}d

^{2}+ 25d

^{4}[From the Binomial Theorem]

5(2d

^{2})

^{2}= -20d

^{4}

Now, (2*5

^{2}r) is a fifth power (#7e) so (2*5

^{2}r)

^{2}is a fifth power [since (x

^{5})

^{2}= (x

^{2})

^{5}]

(2*5

^{2}r)

^{2}= 2*5

^{3}*10r

^{2}= (2*5

^{3})*v = (2*5

^{3})[5d(c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})] =

(2*5

^{4}d)(c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4})

So, we next show that gcd( 2*5

^{4}d ,c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4}) = 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 c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4 }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*5

^{4}d and c

^{4}+ 10c

^{2}d

^{2}+ 5d

^{4 }are 5th powers.

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

c + 5d

^{2}= c'(c'

^{4}+ 50c'

^{2}d'

^{2}+ 125d'

^{4})

2d

^{2}= 5d'(c'

^{4}+ 10c'

^{2}d'

^{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*5

^{8})(2d

^{2}) = 2

^{2}*5

^{8}d

^{2}= (2*5

^{4}d)

^{2}

(12) We know that 2*5

^{4}d is a fifth power from step (#9d) so (2*5

^{4}d)

^{2}is also a fifth power [Because (x

^{5})

^{2}= (x

^{2})

^{5}]

(13) So 2*5

^{9}d'(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4}) is a fifth power by the argument below:

First, we note that (2*5

^{8})(2d

^{2}) is a fifth power since:

(2*5

^{8})(2d

^{2}) = (2*5

^{4}d)

^{2}(from step #11) and (2*5

^{4}d)

^{2}is a fifth power (from step #12)

But this shows that 2*5

^{9}d'(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4}) is a fifth power since:

2*5

^{9}d'(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4}) = (2*5

^{8}) * 5d'(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4}) = (2*5

^{8})(2d

^{2}) [from step #10.]

(14) gcd(2*5

^{9}d' , c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4})=1 since:

(a) Assume there is a prime f that divides both 2*5

^{9}d' and c'

^{4}+ 10c'

^{2}d'

^{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'

^{2}d'

^{2}+ 5d'

^{4}

(c) We can eliminate Case I since is c'

^{4}+ 10c'

^{2}d'

^{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'

^{2}d'

^{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*5

^{9}d' and c'

^{4}+ 10c'

^{2}d'

^{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'

^{2}d'

^{2}+ 5d'

^{4}) = 2d

^{2}[From steps (#10) since c',d' are nonzero and therefore c'

^{2}and d'

^{2}≥ 1]

We note that 5d'(c'

^{4}+ 10c'

^{2}d'

^{2}+ 5d'

^{4}) = 25d'

^{5}+ 5d'(c'

^{4}+ 10c'

^{2}d'

^{2})

Also:

25d'

^{5}≤ 2d

^{2}→ d'

^{5}≤ (2d

^{2})/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:

(2d

^{2})/25 is less than d

^{5}

(ii) Multiplying both sides by 25 gives us:

2d

^{2}is less than 25d

^{5}

(iii) Since d ≥ 1 (#7), we know that d

^{2}is less than d

^{5}.

(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

## 13 comments:

Step (3):

Is there an errant minus at the start of:

-2^5m*5^5n*z'^5 = 2p(p^4 + 10p^2q^2 + 5q^4) ?

Step (6d):

f changes into p and back again.

Rob

Hi Rob,

Thanks for the comments! I have tried to rewrite the proof to make it clearer.

There was an errant minus. It has been removed.

The argument in (6d) is confusing. I have changed it to make it clearer.

Cheers,

-Larry

I think you left out the square on a in 6e, or else i don't follow the proof. Other than that i love the lay out real easy to follow

Thanks

Tim

I think you are missing a square on the a in 6e or else i don't make the conection. Love the proof though makes it real nice and simple to follow thank you

Tim

Hi Tim,

Thanks for your comment. I've updated step #6e. You are right that the detail was not clear.

Please let me know if you have questions about any other steps.

Cheers,

-Larry

I think in step (6a):

Since f divides

a, we know that f divides 10r^2Should be:

Since f divides

b, we know that f divides 10r^2Rob

In step (9d) should

Now, (2*5^2r) is a fifth power (#7e)be

Now, (2*52r) is a fifth power(#6e)Rob

In step (9d) should:

(2*5^2r)^2 = 2*5^3*10r^2 = (2*5^3)*vbe

(2*5^2r)^2 = 2*5^3*10r^2 = (2*5^3)*bRob

In step (9d) should:

(d) u' - 5v'2 is a fifth power since:be

(d) u'^2- 5v'2 is a fifth power since:Rob

In step (10):

(10) With the properties in step #9, we can use a lemmawhich I will prove laterto show:Is the proof of this lemma up yet?

:-(

Rob

Apologies for the last post.

Step (10) is similar to step (7) and uses the same lemma.

Right?

Rob

In step (6e) should

(e) t=a - 5b^2 is a fifth power since:be

(e) t=a^2- 5b^2 is a fifth power since:as in step (5d)?

Rob

In step (8) should:

u' = c + 5d^2be

u' = c^2+ 5d^2Rob

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