# Is this a proof of Fermat's Last Theorem where n = 3?

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.

• How do you justify the sentence “This means $\frac{p}{q}$ is actually a cube over a cube”? – José Carlos Santos Jul 22 '18 at 12:24
• The very definition of $x$ implies that $x=\frac{B^3}{A^3}$ – random Jul 22 '18 at 12:27
• "Therefore $\lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$\lambda^3$ cannot be a fraction", then justify why this is the case ! – Peter Jul 22 '18 at 12:30
• Why is "$1+\frac pq$ also a cube"? – Arnaud Mortier Jul 22 '18 at 12:55
• After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+\frac p q A^3 = C^3$ ? – random Jul 22 '18 at 13:14

You have $$qD^3=v E^3.\tag{1}$$ But if you've followed carefully, you'd have noticed that $D=C$, $E=B$, $p=A^3$, $q=B^3$ and $v=C^3$. The conclusion given is then that "$\lambda^3$ doesn't have to be a fraction". I presume this is intended to mean that $D^3/E^3\notin\Bbb Z$. But (1) says that $B^3C^3=C^3B^3$. From this tautology, why can we deduce anything non-trivial at all, let alone $D^3/E^3\notin\Bbb Z$?