Proof of a special case of Fermat's Last Theorem. Here, I will try to prove a special case of Fermat's Last Theorem, namely when $a=b$ in this definition:

Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation $a^n + b^n = c^n$ for any integer value of $n$ greater than two.

Here's the proof:
Let's suppose the equation $x^n+x^n=y^n$ have one or more solutions where $n$ is an integer grater than $2$ and $x$ and $y$ are also integers. This means that $2x^n=y^n$, thus $\sqrt[n]{2}x=y$. But this is a contradiction since $\sqrt[n]{2}$ is irrational for every $n$ in $\Bbb N$, which implies that $y$ is irrational contradicting our second assumption. Therefore, the equation $x^n+x^n=y^n$ has no solution in $\Bbb N$.
Do you think it is right?
Thank you.
 A: On the other hand, you can do the same by noting that $\sqrt[n]{2}$ is a root of 
$$x^n - 2 = 0$$
and rational root theorem says that the only rational roots, if any, must be one of
$$\{-1, 1, -2, 2\}$$
none of which are proper candidates for $n > 1$, so $\sqrt[n]{2}$ is not rational. 
Edit : Further, you can just use Gauss's lemma by noting that 
$$x^n - 2$$
is irreducible over $\mathbb{Z}[x]$, as the sign changes occur in the interval $(2, 1)$  and the negative counterpart assuming for all $n > 1$, none of which contains integers, and thus by Gauss's lemma is also irreducible over $\mathbb{Q}[x]$.
A: To show that $\;\sqrt[n]2\notin\Bbb Q$ let's assume the contrary. Let's assume that two integeres exist $a$ and $b$ (let's take n>2) such that 
$$
\;\sqrt[n]2=a/b
$$
with $a/b$ an irreducible fraction. Now is easy to see that $a$ must be even since
$$
a^n=2b^n
$$
therefore there must exist an integer $k$ such that $a=2k$ and therefore
$$
2^nk^n=2b^n
$$
so it follows that
$$
b^n=2^{n-1}k^n
$$
and since $n>2$ also b must even. But that contradicts the hypothesis that $a/b$ si an irreducible fraction. QED
A: This question is old, but I think a little bit more general proof is :
let $f(x)$ denotes the number of prime factors of $x$. Then $f(a^n)=0 (\text{mod }n)$. If $n > 1$ and $p$ is a prime then $f(p(a^n))= 1 (\text{mod }n)$. So there is no solutions to $p(a^n) = c^n$. 
