Proof of special case of Fermat's Last Theorem Please help evaluate mathematical clarity of a proof of this simple proposition:
If $n \geq 3$
is an odd integer and
$x, y, z > 0$
are integers such that
$x^{n} + y^{n} = z^{n},$
then
$z$
cannot be a prime-power.
Proof.
To prove the assertion note for a start that if there exists a counterexample to it
then we have
$$z < x+y < 2z,$$
where the inequality on the left follows from the relations
$(x+y)^{n} > x^{n} + y^{n} = z^{n}$
and the inequality on the right follows from the relations
$x, y < z$
which follow from the implications
$x \geq z \Rightarrow y \leq 0$
and
$y \geq z \Rightarrow x \leq 0.$
Moreover,
from the equation
$x^{n} + y^{n} = z^{n}$
and that
$n$
is an odd integer,
it follows that
$$(x+y) \mid z^{n}.$$
Since
$z$
is a prime-power,
there exist a prime
$u$
and a positive integer
$m$
such that
$$z^{n} = (u^{m})^{n}.$$
By the inequality
$z < x+y$
there is a positive integer
$m'$
such that
$$x+y = u^{m'},\ \ m+1 \leq m' \leq mn,$$
whence
$$x+y = u^{m'} \geq 2u^{m} = 2z$$
(with equality if
$m = 1$
and
$u = m' = 2$),
contradicting the inequality
$x+y < 2z.$
Thus $z$
cannot be any prime-power.
 A: Your statement is false. If $n=1$, then $$1^1+1^1=2^1$$ shows that $x+y$ can be a prime, and $z$ can be a prime power.
The problem in your proof is that you say $z<x+y$ which is not true in this case since $(x+y)^n$ is not greater than $x^n + y^n$.
AFTER EDIT:
I find your proof very hard to read. My complaints are:


*

*In the beginning, you don not specify explicitly that you are proving by contradiction. You only say that if a counterexample exists, then something something. A more readable way would be to say 



Let us prove the statement by contradiction. Let $x,y,z$ be such numbers that $x+y$ is prime or $y$ is a prime power

Now, my MAIN objection: in the middle of your proof, you say "if $z$ is a prime power". Sure, I agree, but even in a counterexample you assume exists for the purpose of your proof by contradiction, the number $z$ may not be a prime power. Even simply having $x+y$ be a prime is enough to constitute a counterexample.
After your second edit:
My initial complaint still stands. In the beginning, it is unclear that you are using contradiction.Apart from that, you write that from $x^n + y^n = z^n$ and $n$ being odd, it follows that $(x+y)|z^n$ which could use some clarification (just to make it clear).
The next unclear point is the inequality $m+1\leq m' \leq mn$. Where did this come from? Sure, from $u^m=z<x+y=u^{m'}$, I can see that $m'\geq m+1$, but why is $m'\leq mn$? You need to clarify this in your proof.
