Prove $x^n+y^n=z^n$ has no solution in natural numbers for $n\geq z$ If $ x$, $ y$, $ z$, and $ n$ are natural numbers, and $ n\geq z$ then prove that the relation $ x^n + y^n = z^n$ does not hold.
I have a question: Can I use Fermats Last Theorem here directly? It makes my life easier, but I have a doubt whether or not in Olympiads (if at all they ask such a question; most probably they would not :D) I can directly state :By Fermats Last Theorem... and then proceed, for till now I dont think I have come across a simple and elegant proof for the theorem. I would like to have suggestions on this. Please help. Any other solution is also welcome.
 A: Here’s a proof without using FLT. Without loss of generality assume $x \leq y$. Clearly $z>y$, so $n \geq z \geq y+1$. Then $$z^n \geq (1+y)^n \geq y^n+ny^{n-1} \geq y^n+(1+y)y^{n-1}>2y^n \geq x^n+y^n$$
A: Mathematically, using Fermat's Last Theorem would be a valid proof. Of course, you need to consider the case $z=1$ and $z=2$ separately, because FLT needs $n > 2$, and you have $n > z$. If they were to ever give this problem in IMO, an FLT solution should receive full points (and that's why they'd never give such a problem there).
Of course, if this is a practice problem, and your goal is to learn number theory rather than produce a proof, you probably should only use theorems that you can prove.
A: I think the object of the question is to prove that $x,y,z,n\in \mathbb N \wedge x^n+y^n=z^n\Rightarrow z\ge 3$, and hence $n\ge 3$. Having proven this much, invoking FLT would be appropriate. 
Also, the question must be using the (general but not universal) convention $0\not \in \mathbb N$ since $0^n+0^n=0^n$ for all $n\ge 1$, making $n>z=0$. 
So if $x,y \ge 1$ then $x^n+y^n\ge 2$ requiring $z\ge 2$. Although $1^1+1^1=2^1$, this does not satisfy $n\ge z$. Also, $1^2+1^2\ne 2^2$. If $x>1$ or $y>1$, then $z>2 \Rightarrow n\ge 3$.
