I found this blog which includes many proofs that related to Fermat's Last Theorem (and some other proofs). These proofs are not the original ones from Fermat or Euler, but they are somewhat 'fun' to read through. (instead of z squared, the blog used z to the fourth)

I also found this PDF, which is proving the n=4 case in Fermat's Last Theorem by using the Pythagorean Triples and the method of infinite descent.

Lin

I took Abstract Algebra last semester and even though we did not spent much time exploring the use of Fermat's theorem, I would like to point out the importance of the theorem in the subject. In class, we learned that given Zp (p stands for primes), the elements 1,2,3,…, p-1 form a group of order p - 1 under multiplication modulo p. From this, we arrived at the Fermat's theorem where it's application makes it very easy to compute

the remainder of 8^103 divided by 13. The theorem is well used throughout Mathematics and that is how I got to use Fermat's theorem.

Lenim Depina

Since we're skipping Sophie Germain I looked into some of her work to see if it would be interesting. It seems that she did a lot of work with Fermat's last Theorem and prime numbers. I found a biography describing some of her work. The part most relevant to our class I think is her "grand plan" to prove Fermat's theorem and how this led to a discovery of a new form for prime numbers 2p+1 if both 2 and p are prime.

Theorem.http://www.agnesscott.edu/lriddle/women/germain-flt/sgandflt.htm

-Angela

Euclid was so smart, I really want to know how could he get the formula for the perfect number,

"if for some k>1, 2^{y}-1 is prime then 2^{k-1}(2^{k}-1) is a perfect number"

So, I need to how could he prove it.

Proof,

Suppose 2^{n}-1 is prime, let a=2^{n-1}(2^{n}-1). Then, n>=2, which means 2^{n-1} is even and hence so is a=2^{n-1}(2^{n}-1)..

Note that 2^{n}-1 is odd. Since all divisors(except 1) of 2^{n-1} are even it follows that 2^{n-1} and 2^{n}-1 are coprime.

let &(n) be the sigma function of n, that is, the sum of all divisors of n, including n.

from Signma function is Multiplicative , it follows that &(a)=&(2^{n-1})&(2^{n}-1).

But as 2^{n}-1 is prime, &(2^{n}-1)=2^{n} from Sigma of Prime Number,——(let &(n)be the sigma function of n, then

&(n)=n+1 if and only if n is prime.. PS : I do not know to get this Theorem..)!!

Then we have that &(2^{n-1})=2^{n}-1 from Sigma of power of prime. let&(n) be the sigma function of n, that is, let &(n) be the sum of all positive divisors of n, then &(n)=P^{k+1}-1 / p-1

Hence it follows that &(a)=(2^{n-1})2^{n}=2a.

Hence from the defintion of perfect number it follows that 2^{n-1}(2^{n}-1) is perfect.

PS: still confused for a lot, like the Sigma function Theorem.. what's the formula and how does it come from

Jun