bassam king karzeddin

2007-02-21 14:51:51 UTC

We have the following general equation (using the general binomial theorem)

(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p

Where

N (x, y, z) is integer function in terms of (x, y, z)

P is odd prime number

(x, y, z) are three (none zero) co prime integers?

Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)

(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)

CASE-1

If (p=3) implies N (x, y, z) = 1, so we have

(x+y+z)^3 =3* (x+y)*(x+z)*(y+z)

Assuming (3) does not divide (x*y*z), then it does not divide (x+y)*(x+z)*(y+z),

So the above equation does not have solution

(That is by dividing both sides by 3, you get 9 times an integer equal to an integer which is not divisible by 3, which of course is impossible

I think proof is completed for (p=3, and 3 is not a factor of (x*y*z)

My question to the specialist, is my proof a new one, more over I will not feel strange if this was known few centuries back

Thanking you a lot

Bassam King Karzeddin

Al-Hussein Bin Talal University

JORDAN