Can someone point out the error in the following "proof": Prove a^n + b^n =/ c^n for n>2, a,b,c>1 (=/ means not equal to) Let b=xa where x>1 and is from the set of real numbers generated by fractions, such that b is an integer so: a^n + (xa)^n =/ c^n Expanding a^n + x^n.a^n =/ c^n then, taking the common factor a^n out a^n(1+x^n) =/ c^n then dividing through by a^n 1+x^n =/ c^n/a^n Substitute y from the set of real numbers given by the fraction c/a, then 1+x^n =/ y^n or: y^n - x^n =/ 1 Factorizing: x^n - y^n = (x - y)*(x^[n-1] + x^[n-2]*y + ... + y^[n-1]). If the left side equals 1, then x > y, and x - y must be a positive divisor of 1, namely 1. Then x = y + 1. Substitute that into the second factor above, and set that second factor also equal to 1. That should give you a contradiction. The contradiction means that the assumption that the left side equals 1 must be false, and then you're done.