visual diagram of proof

This is the section of number line where x,y, z, x+y located relative to each other (order is important, not distance)

_______ X ________ Y ___ Z ___ X + Y ____

Here is section of number line where X^2 to (X+Y)^2, with Y^2, Z^2, (X^2 +Y^2), etc


___ X^2 ___Y^2__ (X^n + Y^n)^2/n __ (X^2 + Y^2)^2/2


Now Z^2 =(X^n + Y^n)^2/n and is less than (X^2 + Y^2)^2/2 =(X^2 + Y^2)^1

So Z^2 < (X^2 + Y^2)^1

There must exist a value, call it a, where 0 < a < 1 where

Z^2 = (X^2 + Y^2)^a

But if X,Y, and Z are whole numbers, then the left hand side of equation is a whole number, while the right hand side is an non-whole number.
This is true because a whole number raised to a non-whole number is an non-number number.

Thus this contradiction proves there is no X,Y,Z all being whole numbers that can meet this requirement.

another counter-example

Let's look at the familiar equation 3^2 + 4^2 = 5^2

What happens to 3^3 + 4^3 = 73 ? The cube root of 73 is 4.17 (approximately)

What happens to 3^4 + 4^4 = 291 ? The 4th root is 4.13 (approximately.

The progression of the higher roots is toward the number 4.
5, 4.27, 4.13 As the root gets higher in number, the root gets closer to 4. While the square root may be greater than y + 1, eventually the distance between z and y is less than 1. this means that there is a finite number of possible solutions for z, when x,y are fixed. So there are only a finite number of steps to consider for each z.

some counter examples

Here is one example that shows part of the process.

I claim x,y are 2 different numbers.
suppose x=y then X^n + X^n = Z^n

or 2X^n= Z^n

This implies that either x or Z must have a factor of 2^1/n as part of the number. One of them is irrational.