# Attempt to prove Fermat's Last Theorem n = 3. [duplicate]

My previous attempt at proof did not work as I thought and so I decided to try again. Please bear in mind that this is not a duplicate of my old proof. If you look at the other one you will see that this examines the parity of the equation, something the previous attempt does not do. This time I have based my attempt on induction rather than algebra.

To begin with, I examined the equation: $$A^3 + B^3 = C^3$$ Where $A$, $B$, $C$ $\in \Bbb Z$ and $A$, $B$, $C > 0$. We can also assume that $A, B, C$ are the lowest solutions. Therefore, $A^3, B^3, C^3$ are co-prime.

Examining this, we should be able to draw conclusions on each term's parity. In this form, we could have: $$(1)EVEN + ODD = ODD$$ $$(2)ODD + ODD = EVEN$$ $$(3)EVEN + EVEN = EVEN$$

However, equation $3$ is not possible as we know that $A^3, B^3, C^3$ are the lowest solutions possible and are co-prime. Therefore, we are left only with $1$ and $2$.

My previous attempted solution involved rewriting the equation in terms of $A^3$: $$A^3 + xA^3 = C^3$$ Here $x$ is either an integer, irrational or fraction. If it is an integer we get: $$(1+x)A^3 = C^3$$ Since $x \in \Bbb Z$, $x+1 \in \Bbb Z$. Therefore, $\sqrt[3]{x+1} \in \Bbb Z$ and $\sqrt[3]{x} \in \Bbb Z$. The only possible integer solution to this is $x = 0$. However, this would mean that $B^3 = 0$. Therefore, $x$ is irrational or a fraction. If $x$ is irrational, $B^3$ will be irrational.

Therefore, $x$ is a fraction.

We can now represent this in the equation: $$A^3 + \frac{p^3}{q^3}A^3 = C^3$$ Where $p,q \in \Bbb Z$. The reason the fraction can be represented as such is because that is the ratio of $A^3$ to $B^3$. Even if $p^3$ and $q^3$ did share factors they could be simplified to still be cubes. (For example, $\frac{64}{8} = 8$.)

Simplifying this further we get: $$(qA)^3 + (pA)^3 = (qC)^3$$

We shall begin by analysing $qA$. We know from earlier that this is either odd or even. If we assume it is even, then either: $$\frac{A}{2} \in \Bbb Z , \frac{q}{2} \notin \Bbb Z$$ or $$\frac{A}{2} \notin \Bbb Z , \frac{q}{2} \in \Bbb Z$$ or $$\frac{A}{2} \in \Bbb Z , \frac{q}{2} \in \Bbb Z$$

Analysing $pA$, we can see from before that $pA$ must be odd. Therefore: $$\frac{A}{2} \notin \Bbb Z , \frac{p}{2} \notin \Bbb Z$$ This means we can compare this with the other possible parities of $A$ to give:

$$\frac{A}{2} \notin \Bbb Z$$ $$\frac{p}{2} \notin \Bbb Z$$ $$\frac{q}{2} \in \Bbb Z$$

Substituting this all into our equation gives us: $$(EVEN)^3 + (ODD)^3 = (EVEN)^3$$ $$EVEN + ODD = EVEN$$

This clearly is incorrect and so our original assumption, that there are integer solutions to the below is incorrect. $$A^3 + B^3 = C^3$$

Q.E.D. (Hopefully.) Is this correct? I cannot find anything wrong with it unlike my previous one. Thank you.

• You forgot about ODD+EVEN=ODD (and in fact the writing $(qA)^3 + (pA)^3 = (qC)^3$ can only be of that form). – Arnaud Mortier Jul 22 '18 at 20:57
• You've already asked this question. and you accepted an answer to that same question. Do not re-post a question you've already posted. Instead, edit the original. – amWhy Jul 22 '18 at 21:00
• @amWhy I think this is not a duplicate, because the proof attempt is quite different. – AlgebraicsAnonymous Jul 22 '18 at 21:10
• @amWhy This proof attempt seems very different from the other one; why shouldn't it deserve it's own question? – Sambo Jul 22 '18 at 21:29
• The observation made by @ArnaudMortier seems on point. You begin the second half of the proof with "If we assume it [$qA$] is even, then..." But you never follow up with "If, on the other hand, $qA$ is odd, then..." You need to consider that case. – Brian Tung Jul 22 '18 at 21:44