TITLE: Prove that the relation $x^n + y^n = z^n$ does not hold for $n \geq z$
QUESTION [5 upvotes]: Assume $x, y, z$, and $n$ are positive integers, and $n \geq z$. Prove that the relation $x^n + y^n = z^n$ does not hold.

I find it hard to relate the condition $n \geq z$ to solve this question. Maybe this is best proved by induction. I could start with $n = 1$. Then we have $x+y = 1$ for which there exist no positive integer solutions $(x,y)$. We can then assume the statement is true for some $k \geq z$. Thus there exist no positive integer solution pairs $(x,y)$ to $x^k+y^k = z^k$. We have to show it is the same for $x^{k+1}+y^{k+1} = z^{k+1}$.

REPLY [7 votes]: Since we assume that $x,y,z$ are positive integers, we have $x^n + y^n > y^n$, and since $a \mapsto a^n$ is monotonic, it follows that $x^n + y^n > z^n$ if $y \geqslant z$. By the same reasoning we can exclude $x \geqslant z$.
For $x,y < z$, we then have $x^n + y^n \leqslant 2(z-1)^n$, and then showing $2(z-1)^n < z^n$ finishes the proof. Since by assumption $n \geqslant z$, we have
$$\frac{(z-1)^n}{z^n} = \biggl(1 - \frac{1}{z}\biggr)^n \leqslant \biggl( 1 - \frac{1}{z}\biggr)^z,$$
and it is easily shown if not already known that
$$\biggl(1 - \frac{1}{z}\biggr)^z < e^{-1}$$
for all positive integers $z$. Hence
$$\frac{2(z-1)^n}{z^n} < \frac{2}{e} < 1.$$