Daniel Niv proof 1

Initial Comments [I intend on reading proof in entirety when I remember all my number theory]:

Ok..got lost on part 4. Here are my comments on the first 4 parts, I will comment on rest once you clarify:

(1) x^n can be computed from the generating formula I showed you earlier during camp x^n= x[s(n,1)]+ x(x-1)[s(n,2)]...+x(x-1)...(x- n+1)[s(n,n)]. Thus I express x^n + y^n - z^n as x[s(n,1]... + y[s(n,1]... - z[s(n,1)]...

Good

(2) Which can be rewritten as s(n,1)[x+y-z]+ s(n,2)[x(x-1)+y(y-1)-z(z-1)]... s(n,n)[x(x-1)...(x-n+1)+y(y-1)...(y-n+1)-z(z-1)...(z-n+1)].

Good, just regrouping

(3) In order to write this more easily I represent x(x-1)...(x-r+1)+y(y-1)...(y-r+1)-z(z-1)...(z-r+1) as M sub r, here denoted Mr, since I can't write it correctly with hotmail. And I represent s(n,k) as S sub k or in this case Sk. Thus, x^n+y^n-z^n= S1M1+S2M2+S3M3...SnMn.

For subscript, do M_r. Ok, so new notation for the sum of products. Good.

(4) In other words, since x^n+y^n-z^n=0; 0=S1M1+S2M2+S3M3...SnMn. Now, here is my idea. From number theory class we learned some nice facts. For example r!/ x(x-1)(x-2)...(x-r+1) since I am multiplying n numbers. Similarly r!/ y(y-1)...(y-r+1) and r!/ z(z-1)...(z-r+1). From this information, since I know that S1...Sn are integers r!/ Sr[Mr] and r!/ Sr+1[Mr+1] and all k such that n>=k>r because r! / x(x-1)(x-2)...(x-r) this is x(x-1)(x-2)...(x-r)[x-r+1].

Now here is where I lost you: r!/ x(x-1)(x-2)...(x-r+1) is not a statement, but a number. Do you mean divides i.e r!| x(x-1)(x-2)...(x-r+1)???

Comments: Follow-up Sure but could you use the fixed version which I descibed to you; in which, my result is that (p+1)! divides x^p+y^p-z^p where p is prime iff p-1/n-1. Because we assumed x^n+y^n-z^n=0. Yes, I would definately enjoy you putting the proof up there. Only that instead of saying this works for all primes p, it only works for primes where p-1/n-1. And thank you for writing me to put up my work. This makes me happy.- Math ninja apprentice #2 out.A.K.A. Daniel Niv

Also, before I left while I was doing homework I got this really nice idea for work on Fermat's Last Theorem. The reason I am looking at this is because this has to do with a semi-new type of mathematics/ notation that I have been using. This actually seems to have a lot of nice generating formulas and has to do with Stirling's triangle among other things. Except I just wanted to tell you the basic outline of my idea...

x^n can be computed from the generating formula I showed you earlier during camp x^n= x[s(n,1)]+ x(x-1)[s(n,2)]...+x(x-1)...(x- n+1)[s(n,n)]. Thus I express x^n + y^n - z^n as x[s(n,1]... + y[s(n,1]... - z[s(n,1)]... Which can be rewritten as s(n,1)[x+y-z]+ s(n,2)[x(x-1)+y(y-1)-z(z-1)]... s(n,n)[x(x-1)...(x-n+1)+y(y-1)...(y-n+1)-z(z-1)...(z-n+1)]. In order to write this more easily I represent x(x-1)...(x-r+1)+y(y-1)...(y-r+1)-z(z-1)...(z-r+1) as M sub r, here denoted Mr, since I can't write it correctly with hotmail. And I represent s(n,k) as S sub k or in this case Sk. Thus, x^n+y^n-z^n= S1M1+S2M2+S3M3...SnMn. In other words, since x^n+y^n-z^n=0; 0=S1M1+S2M2+S3M3...SnMn. Now, here is my idea. From number theory class we learned some nice facts. For example r!/ x(x-1)(x-2)...(x-r+1) since I am multiplying n numbers. Similarly r!/ y(y-1)...(y-r+1) and r!/ z(z-1)...(z-r+1). From this information, since I know that S1...Sn are integers r!/ Sr[Mr] and r!/ Sr+1[Mr+1] and all k such that n>=k>r because r! / x(x-1)(x-2)...(x-r) this is x(x-1)(x-2)...(x-r)[x-r+1]. This number is a multiple of x(x-1)...(x-r+1) which means it is divisible by r!. Here is my strategy. I plug in 3 for r and I get 3!/ S3M3+S4M4...SnMn; I also know 3!/ 0, the other side of the equation. Thus 3!, or 6, must divide S1M1 + S2M2. Sadly. If you fiddle around with this we don't learn any new information. However, when r=4; we get 24/ S1M1+S2M2+S3M3 or S1(x+y-z) + S2(x^2+y^2-z^2 - (x+y-z)) + S3( x^3+ y^3 - z^3 -3( x^2 + y^2 - z^2) + 2(x+y-z)). Now from the generating formula for stirling's triangle for the nth row. S1= 1^n-0^n; S2=[(2^n-2(1^n)+0^n)/2]; S3= [(3^n-3(2^n)+3(1^n)-0^n)/6]. After substituting in these values for S1, S2, and S3 after a small amount of work we can achieve our desired result. First, let me explain that x^r+y^r-z^r is defined as Z sub r, here denoted Zr. Thus, after reducing the Z1's we find that 24/ the coefficient of Z1. Similarly we find that 24/ the coefficient of the reduced Z2's. And finally we get that 24/ (S3)(Z3). By looking at S3 and by using fermat's little theorem, we can see after plugging in some numbers and remembering the properties of primes that 6/ 3^n-3(2^n)+3(1^n)-0^n; however, this information is trivial because to get S3 we divide out 6 on the bottom. Thus, when we look to see if (6)(3)/ 3^n-3(2^n)+3(1^n)-0^n we see that 9 doesn't divide 3^n-3(2^n)+3(1^n)-0^n. Thus, three divides Z3 and we look to see if (6)(8) / " " same as the line before. Except to save us some trouble we look to see if (6)(2)/ 3^n-3(2^n)+3(1^n)-0^n. However; 4 does not divide 3^n-3(2^n)+3(1^n)-0^n. Thus, (3^n-3(2^n)+3(1^n)-0^n)/6 is an odd number and is not divisible by two. Thus, strangely enough 24/ x^3+y^3-z^3. In order words, 4!/x^3+y^3-z^3. By looking at the general case, I found if p is a prime and n>p>2; then (p+1)!/ x^p+ y^p -z^p. Iff X^n + y^n -z^n =0. Thus p/ x^p + y^p - z^p. And by fermat's little theorem. x^p is congruent to x mod p; thus p/x+y-z. Thus: 2,3,5,7,...,n/(x+y-z) and (p+1)!/ (x^p+y^p-z^p). This is an amazing result because it shows that if Fermat's last theorem holds for any n, n and all the primes up to n/ (x+y-z). Also, this gives some interesting properties for x^p+y^p-z^p. Please give me any comments about any mistakes you see in my argument. I worked on this for a short while previously because I was struck by this idea. This seems to work and I would like to know if my argument is correct. Because if so, I have done what Sophie Germain was trying to do in order to make x+y-z very large. I have more work that I can do with this one statement but I was just wondering if you agreed with the parts of this proof except for the general case which is similarly outlined but which I have not included. Thank you very much for your opinion/suggestions. I feel like this might just be a really good original proof. The reason I am interested in the mathematics is that even though this is used to prove a result concerning Fermat's Last Theorem, I can use similar approaches dealing with other problems with exponents. I feel like this is a very useful method and may become very useful. Thank you. This is Daniel Niv, out.