Wednesday, October 23, 1996
Stating the obvious. More on this FLT thing
I pretty much realize that it's probably well known and obvious that with FLT, neither x,y nor z can be divisible by n or there's a nice contradiction. It was fun proving it to myself anyway.
Here's something else that I haven't read but I guess is well established. I call it a strong case of Fermat's Little Theorem.
Given a-b divisible by n, a^n - b^n must be divisible by n^2
So Fermat's Last Theorem can be written as
Given x,y,z relative prime; n odd prime
no solutions exist unless (x+y)^n - (x^n + y^n) is divisible by n^2
Wonderfully obvious but new to me. Now I've noticed that x and y can be written in terms of n like x=an+f and y=bn+g. If (x+y)^n - (x^n + y^n) were divisible by n^2 this would require that f^n + g^n equal some h^n. Seems there's some infinite regression sort of thing.
Notice that for n=5 the above equals 5xy(x+y)(x^2 + xy + y^2) which would mean that x^2 + xy + y^2 must be divisible by n. It's easy enough to see that it can't be.
Here's something else that I haven't read but I guess is well established. I call it a strong case of Fermat's Little Theorem.
Given a-b divisible by n, a^n - b^n must be divisible by n^2
So Fermat's Last Theorem can be written as
Given x,y,z relative prime; n odd prime
no solutions exist unless (x+y)^n - (x^n + y^n) is divisible by n^2
Wonderfully obvious but new to me. Now I've noticed that x and y can be written in terms of n like x=an+f and y=bn+g. If (x+y)^n - (x^n + y^n) were divisible by n^2 this would require that f^n + g^n equal some h^n. Seems there's some infinite regression sort of thing.
Notice that for n=5 the above equals 5xy(x+y)(x^2 + xy + y^2) which would mean that x^2 + xy + y^2 must be divisible by n. It's easy enough to see that it can't be.