John Grieve: Fermat's Last Theorem X^n Y^n = Z^n    
 Fermat's Last Theorem X^n Y^n = Z^n0 comments
1 Aug 2011 @ 10:21, by John Grieve


I believe that the essence of this problem is contained in the simple geometrical fact that the six sides of a cube are all squares. Thus in the equation X^3 + Y^3 = Z^3 there are also squares X^2, Y^2 and Z^2 but it is not the case here that X^2 +Y^2=Z^2, that is X^2 plus Y^2 is not equal z^2 since this would involve a contradiction.

But there are still definite constraints on the values which X,Y and Z can take as they are both the sides of the cubes and the squares. So if X^3 + Y^3 = Z^3 then we would find values of the squared equation close to but not the same as the cubed.

We see from the following Pythagorean Triplets that as the value of n goes higher the gaps get smaller;
6 plus 8 equals 14
6^2 plus 8^2 equals 10^2
6^3 plus 8^3 equals 8.xxxxxxxx^3
6^4 plus 8^4 equals 8.xxxxxxxxxxxxxxx^4

We notice from this that after the squared equation the values of Y and Z get ever closer, which suggests that Z can't be an integer, in fact the limit of its value is the lower number 8.

This suggests why X^3 +Y^3=Z^3 is an impossible equation for integer values of the variables. The sides of the cubes are also the squares and the two share the same values of X,Y and Z.


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