<p>Ya true. lol. However, the power to which the expression is raised is (n/2). Thus, the non-integer can only be of the form a^(2/n), which leaves us only with finite possibilities for a particular n.
Nevertheless, I’ve found a flaw in my proof…it kinda baffles me that I did not find it before.
c^2=a^2 + b^2-2abcosC. Hence, c^n=[a^2+b^2-2abcosC]^(n/2). First I have equated, and then I say that this is not true. It’s like saying a=b, and then saying a is not equal to b.
The statement that I have to actually prove is [a^2+b^2-2abcosC]^(n/2) cannot be equal to x^n+y^n for any integers a, b, x and y. Moreover, a and b can be assumed to be relatvely prime, and so can y and z.
As you might have guessed, I have acquired some knowledge of number theory. :)</p>