So I have spent some time thinking about Fermat's Last Theorem and about how to come up with a proof for certain cases of n. To begin with I took n = 3.
$$A^3 + B^3 = C^3 (A,B,C \in \mathbb {Z})( A,B,C ≠ 0)$$ We can assume that $A<B<C$. Because of this we can rewrite this as: $$A^3 + xA^3 = C^3$$ We now have three possibilities: $x$ is an integer, $x$ is an irrational, or $x$ is a fraction.
If $x$ is an integer then $x$+1 and $x$ must be cubes. The only numbers which satisfy this are $x$ = 0. However, this means $B^3 = 0$.
$x$ cannot be irrational because then $C^3$ is irrational. Therefore $x$ is a fraction.
We can represent this now as: $$A^3 + \frac{p}{q}B^3 = C^3 (p,q \in \mathbb {Z})(GCD(p,q) = 1)$$ This means $\frac{p}{q}$ is actually a cube over a cube. Therefore $p$ and $q$ are cube numbers.
However, $1 + \frac{p}{q}$ is also a cube. Therefore $\frac{p+q}{q}$ is a cube over a cube. Hence: $$\sqrt[3]{p} ,\sqrt[3]{q}, \sqrt[3]{p+q} \in \mathbb {Z}$$
We can now create another variable $v$ for $p+q$. Therefore, $$p+q = v$$ (Where $\sqrt[3]{v} \in \mathbb {Z}$)
Now we must deal with $\frac{p}{q}+1$. Since this is a cube we can write this as: $$\frac{p}{q} + 1 = \lambda^3$$ Since $p$ and $q$ are co-prime and in simplest form, $\lambda$ and $\lambda^3$ are also fractions in simplest form. Rearranging, we get: $$p = q(\lambda^3 - 1)$$
Substituting this into $p+q=v$, we get: $$q+q(\lambda^3 - 1) = v$$ $$q\lambda^3 = v$$
Since $\lambda^3$ is a fraction, we can express it as $\frac{D^3}{E^3}$. Hence: $$qD^3 = vE^3$$
However, because $v$ is a cube, $vE^3$ is also a cube. Therefore $\lambda^3$ doesn't have to be a fraction. This is a contradiction, and so there can be no integer solutions for $p$ and $q$. Subsequently, there can be no integer solutions for A,B,C in: $$A^3 + B^3 = C^3$$.
If anyone could point out the problem in this and what you think of it it would be greatly appreciated. Thank you.
