This argument puzzled me, but unfortunately this is as far as I got: ~~Lemma 1~~: z^n can be written as a difference of squares: Proof: z^n = x^2 - y^2

This argument puzzled me, but unfortunately this is as far as I got: ~~Lemma 1~~: z^n can be written as a difference of squares: Proof: z^n = x^2 - y^2 = (x+y)(x-y), leading to the system of equations x + y = z^(n-1) x - y = z which may be solved for x and y: x = (z^(n-1) + z)/2 y = x - z QED. ~~Lemma 2~~: Any number squared can be written as a sum of sequantial odd numbers starting at 1. By induction: (n)^2 = (n-1)^2 + (2(n-1)+1) = \sum_{k=0}^{n-1}(2k + 1) QED. ~~~Fermat's Last Theorem~~~: z^n = x^n + y^n has no solutions for n>2, and positive z, x, y. Proof: Without loss of generality, assume a > b, then rewrite z^n as a sum of sequential odd numbers starting at b: z^n = a^2 - b^2 = \sum_{k=b}^{a-1}(2k+1) x^n and y^n can similarly be written as a sum of sequenatal odd numbers: x^n = c^2 - d^2 = \sum_{k=d}^{c-1}(2k+1) y^n = e^2 - f^2 = \sum{k=f}^{e-1}(2k+1) By substitution to the Theorem's equation: \sum_{k=b}^{a-1}(2k+1) = \sum_{k=d}^{c-1}(2k+1) + \sum_{f}^{e-1}(2k+1) Which is true if and only if there are no gaps in the bounds of summation on the right side, so d = b, c = f, and e = a. But then a^2 - b^2 = (f^2 + (-a^2))^n + (b^2 + (-f^2))^n and this is certainly not true by the binomial theorem. We've reached a contradiction so QED. ^^ This is where I'm stuck. I don't actually know why that would not be true by the binomial theorem? It seems like simple expansion should do it.