# 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.

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Yes, it is (assuming you can assume $\;\sqrt[n]2\notin\Bbb Q\;$ for $\;2\le n\in\Bbb N\;$ . –  DonAntonio Jan 13 '14 at 11:39
By the way you can easily show that $\;\sqrt[n]2\notin\Bbb Q\;$. I will add an answer since I have more space to write... –  Umberto Jan 13 '14 at 11:58
Al fine, though this is a very special case. –  Hagen von Eitzen Jan 13 '14 at 12:05
Thank you all. :) –  user121011 Jan 13 '14 at 12:06

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

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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 $\mathbb{Q}[x]$.

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Cool. For more information (as a reference) en.wikipedia.org/wiki/Rational_root_theorem –  Umberto Jan 13 '14 at 14:38
I have also added another way to prove this in an edit. See here for reference to Gauss's lemma. –  Balarka Sen Jan 13 '14 at 16:33

This question is old, but I think a little bit more general proof is :

let f(x) = number of prime factors of x.

f(a^n) mod n = 0

if n > 1 and p is a prime f(p(a^n)) mod n = 1

so there is no solutions to p(a^n) = c^n

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