Friday, October 28, 2005

Fermat's Last Theorem: Proof for n=5

For those interested in the history behind this proof, you may want to 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.

Theorem: Proof for FLT: n=5

x5 + y5 = z5 → xyz = 0 if x,y,z are integers.


(1) Assume there is a solution where xyz ≠ 0

(2) From previous results, we know that we can assume that gcd(x,y,z)= 1 (see here).

(3) We can also assume that x,y are odd and z is even since:

(a) We know that at least two of the values are odd since gcd(x,y,z)=1 tells that only 1 at most can be even.

(b) If two are odd, then the third is even since odd + odd = even and odd - odd = even.

(c) If z is even we are good so let's assume that x is even.

We know that there exists z', x' such that z' = -z and x' = -x so x' is even:

(-1)5(x')5 + y5 = (-1)5(z')5

First, we add (z')5 to both sides to give us:

(-1)5(x')5 + y5 + (z')5 = 0.

Then, we add (x')5 to both sides to give us:

y5 + (z')5 = (x')5

Even in this case, we have derived a form z5 = x5 + y5 where z is even.

(4) We can assume that 5 divides xyz from Sophie's proof.

(5) Now, it can be shown that if 5 divides z, then there is no integer solution (see here for proof) and likewise, it can be show if 5 doesn't divide z, then there is no integer solution (see here for proof).

(6) So we have a contradiction and can reject our initial assumption.

QED

4 comments:

Pau said...

Hi Larry. My name is Pau and I'm from Spain. I'm doing 2nd of baccalaureate and also i'm doing a research work called "Fermat's last theorem". I'm doing it with a classmate and your blog is helping us a lot.
Thank you very much, Larry.
A cordial greeting from Spain.

Larry Freeman said...

Hi Pau,

I am very glad to hear that my blog is helping.

Cheers,

-Larry

neat_maths said...

Proof of Case I for n = 5
The following is quite elegant. I have not seen this elsewhere.

The trinomial expansion of (x+y+z)^5 simplifies down to

(x+y+z)^5 = 5(z+x)(z+y)(x+y){x^2+y^2+z^2+xy+xz+yz}

assuming that x^5 + y^5 + z^5 = 0

this expression can be neatly rearranged as

= 5 (z+x)(z+y)(x+y)*
{(x+y)^2+(z+y)^2+(z+x)^2}/2

fascinating isn’t it ?

In mod 5, all non 0 squares are either 1 or 4

(1^2 = 1,2^2 = 4,3^2 = 4,4^2 = 1)

and the sum of 3 squares (of non zero numbers) in mod 5 can never be 0

(they must sum to 1, 2, 3 or 4).

The possible combinations are;
(1+1+1) = 3,
(1+1+4) = 1,
(1+4+4) = 4 and
(4+4+4) = 2

Therefore including (z+x)(z+y)(x+y) which also cannot be divisible by 5 under Case I,

the whole expression on the right hand side can contain only one prime factor of 5.

This makes the left hand side expression impossible as all primes in the expression must be raised to the 5th power, at least and 5 is only present once on the right hand side.

Therefore
(x+y+z)^5 = 5 (z+x)(z+y)(x+y)* {(x+y)^2+(z+y)^2+(z+x)^2}/2

is impossible for case I, where 5 does not divide x, y or z and x^5 + y^5 + z^5 = 0

This completes the proof of Case I for n = 5

Keith Griffiths said...

Hi Larry, concerning:

x5 + y5 = z5 → xyz = 0 if x,y,z are integers

could you expand on what the arrow means.

Thanks
Keith