Question about proof-correctness of FLT for n = 3

Question

Anything wrong with the following?

Theorem:

$\quad$ Let:

$\quad\quad \gcd(x,y,z) = 1$,

$\quad$ Then:

$\quad\quad x^3 = y^3 + z^3$ has no non-zero integer solutions for $(x,y,z)$

Proof:

$\quad$ Assuming $3 \nmid x - y \implies 3 \nmid z$ and making use of the following equation:

$\quad\quad x^3 - y^3 = 3x^2(x - y) - (2x + y)(x - y)^2$

$\quad$ which is verifiable by working out parentheses,

$\quad\quad \implies (x^3 - y^3)/(x - y) = 3x^2 - (2x + y)(x - y)$

$\quad\quad \implies (2x + y)(x - y) \equiv -1 \pmod{3}$

$\quad$ Suppose now $3 \not | y$ and the same holds for $x - z$ we also have:

$\quad\quad \implies (2x + z)(x - z) \equiv -1 \pmod{3}$

$\quad\quad \implies (2x + y)(x - y) \equiv (2x + z)(x - z) \pmod{3}$

$\quad\quad \implies (2x + y)z \equiv (2x + z)y \pmod{3}$

$\quad\quad \implies 2xz \equiv 2xy \pmod{3}$

$\quad\quad \implies y \equiv z \pmod{3}$

$\quad\quad \implies y^3 \equiv z^3 \pmod{9}$

$\quad\quad \implies x^3 \equiv y^3 + z^3 \equiv 2y^3 \pmod{9}$

$\quad\quad \implies (x/y)^3 \equiv 2 \pmod{9}$

$\quad$ which is impossible because of $2$ not being a cubic residue $\pmod{9}$.

$\quad$ So now we know $3$ divides either $y$ or $z$ but not both.

$\quad$ Suppose $3|z \implies 3|x - y$, and note:

$\quad\quad\quad\quad \gcd(x - y,z^3/(x - y))$

$\quad\quad\quad\quad = \gcd(x - y,(x^3 - y^3)/(x - y))$

$\quad\quad\quad\quad = \gcd(x - y,3x^2) = 3$

$\quad$ so we know $9 \nmid x - y$.

$\quad$ Take the following equation:

$\quad\quad (x - y)^3 + 3y(x - y)^2 + 3y^2(x - y) = x^3 - y^3 = z^3 \equiv 0 \pmod{27}$

$\quad\quad \implies 3y^2(x - y) \equiv 0 \pmod{27} \implies 3 | y$

$\quad$ contradicting $\gcd(x,y,z) = 1$

$\quad$ The conclusion is there cannot be any solutions at all.

Answer 1

I shall not edit but answer to my own question and prove only the first case of FLT for $n = 3$ in a more correct way. I have no idea for the second case.

It is easier to take the last equation and prove the first case in the following way:

Theorem:

Let:

$\quad x,y,z$ be none-zero integers

$\quad \gcd(x,y,z) = 1$

$\quad 3 \nmid xyz$

Then:

$\quad x^3 \neq y^3 + z^3$

Proof:

$\quad$Consider:

$\quad\quad x^3 - y^3 = 3y^2(x - y) + 3y(x - y)^2 + (x - y)^3 = z^3$

$\quad \implies \gcd((x^3 - y^3)/(x - y),x - y) = \gcd(x - y,3y^2) = 1$

$\quad \implies x - y = r^3$ and $(x^3 - y^3)/(x - y) = s^3$,

$\quad\quad$for some $r,s$ with $\gcd(r,s) = 1$ and $z = rs$

$\quad$Because of:

$\quad\quad s^3 \equiv (x^3 - y^3)/(x - y) \equiv 1 \pmod{9}$

$\quad$and:

$\quad\quad x - y \equiv r^3 \implies (x - y)^2 \equiv r^6 \equiv 1 \pmod{9}$

$\quad$We have:

$\quad\quad s^3 \equiv 3y^2 + 3y(x - y) + r^6 \pmod{9}$

$\quad \implies 1 \equiv 3y^2 + 3y(x - y) + 1 \pmod{9}$

$\quad \implies 3y^2 + 3y(x - y) \equiv 0 \pmod{9}$

$\quad \implies y^2 + y(x - y) \equiv 0 \pmod{3}$

$\quad \implies yx \equiv 0 \pmod{3}$

$\quad\quad$Which is impossible because of $3 \nmid xyz$.

Answer 2

Claim: Let $\gcd(x,y,z)=1$ and let none of $x,y,z$ be zero. then $x^3\neq y^3+z^3$.

Proof:

(Suppose otherwise that $x^3=y^3+z^3$ has a solution)

Assuming $3\nmid x-y\implies 3\nmid z$...

Assuming $3\mid z$ this implies that $3^3\mid z^3$ implying $3^3\mid x^3-y^3$ and $3^3\mid (x-y)(x^2-xy+y^2)$. This does not directly imply that $3\mid (x-y)$ as it could still be that $3^3\mid (x^2-xy+y^2)$, so your assumption is not necessarily a true assumption. This implies we will need to make more assumptions later on to cover the cases missed by having made these assumptions. Also, this is as good a time to point out as any that your assumption that $3\nmid x-y\implies 3\nmid z$ is the exact same assumption as $3\mid z\implies 3\mid x-y$. They are contrapositives to one another and need not appear as assumptions twice in the same proof.

Further, it is often ineffecient to use implications as assumptions when doing case work. It can be confusing and misleading. The assumption "$3\nmid x-y\implies 3\nmid z$" is logically equivalent to "$3\mid x-y$ or $3\nmid z$." Possibly both, possibly only one or the other.

...Assuming $3\nmid x-y\implies 3\nmid z$ and making use of the following equation $$x^3-y^3=3x^2(x-y)-(2x+y)(x-y)^2$$

which is verifiable by working out parentheses,

$\implies (x^3-y^3)/(x-y)=3x^2-(2x+y)(x-y)$

Division by zero error here. In the case that $x=y$ this is not true. You would have an undefined expression on the left and any number possible on the right. If we did have $x=y$ though, this would imply $z=0$. Still, this should be pointed out by the proof writer, not left as a detail for the reader.

$\implies (2x+y)(x-y)\equiv -1\pmod{3}$

How? In your assumptions, you effectively assumed $3\mid x-y$ or $3\nmid z$. If it was the first, then the above is false as $(2x+y)(x-y)\equiv 0\pmod{3}$ not $-1\pmod{3}$. If it was the second, we would have $3x^2-\frac{x^3-y^3}{x-y}=3x^2-\frac{z^3}{x-y}=(2x+y)(x-y)^2$.

Assuming the second then, $3\nmid z$ implying $3\nmid z^3$, so we learn that $3\nmid x-y$ since the expression on the right is an integer and otherwise the expression on the left wouldn't be. Looking at this in terms of modulo3 we have $-\frac{z^3}{x-y}\equiv (2x+y)(x-y)^2\pmod{3}$. Knowing $3\nmid z$ either $z$ can be $1$ or $-1$ mod3 and similarly, knowing $3\nmid x-y$ then $x-y$ is either $1$ or $-1$ mod3 as well, however no work has been done yet to show how they relate to one another. From the work shown, we can make the claim that $(2x+y)(x-y)\equiv \pm 1\pmod{3}$ but it does not follow from work shown that $(2x+y)(x-y)\equiv \color{red}{-}1\pmod{3}$

Suppose now $3\nmid y$ and the same holds for $x-z$ we also have:

What? "and the same holds for $x-z$"... so we are currently working in the very restrictive set of assumptions that $(3\mid x-y$ or $3\nmid z)$ and $(3\nmid y)$ and $(3\nmid x-z)$? We were missing some cases already for the main set of case work, now we're missing subcases within a case.

$\implies (2x+z)(x-z)\equiv -1\pmod{3}$

As you wrote them, your hypotheses are not symmetrical, as the first set of hypotheses allowed for $3\mid x-y$, but your most recent added hypotheses you do not allow $3\mid x-z$. As we saw before, if you did allow $3\mid x-z$ we would have not been able to reach $(2x+z)(x-z)\equiv \pm 1\pmod{3}$. As before though, we have not successfully shown that it must be negative one, as it could still be positive one mod3.

By ignoring this, your argument continues:

$\implies (2x+y)(x-y)\equiv (2x+z)(x-z)\pmod{3}$

$\implies (2x+y)z\equiv (2x+z)y\pmod{3}$

I do not follow how this substitution was made. We have done no work in showing that $z\equiv x-y\pmod{3}$ and our hypotheses do not imply this. Removing the part of the first assumption that led to a contradiction (which the contradiction could have been explained by the division by zero error rather than the assumption itself), we would have $3\nmid x-y$ and $3\nmid z$ simultaneously, but that doesn't mean that $x-y\equiv z\pmod{3}$. We could have $x-y\equiv 1\not\equiv -1\equiv z\pmod{3}$ or vice versa.

$\implies 2xz\equiv 2xy\pmod{3}$

$\implies y\equiv z\pmod{3}$

"Dividing" by two is fine here because two is not a zero in mod3 arithmetic, but division by $x$ is not allowed for fear of division by zero errors. The correct result here would be $3\mid x$ or $y\equiv z\pmod{3}$. You have not done work to imply that $3\nmid x$ to discount this case and this was not included in any of your hypotheses so far.

Ignoring this again for now and continuing...

$\implies y^3\equiv z^3\pmod{9}$

$\implies x^3\equiv y^3+z^3\equiv 2y^3\pmod{9}$

$\implies (x/y)^3\equiv 2\pmod{9}$

This step of "dividing" by $y$ is only okay because our assumptions included $3\nmid y$. It is incorrect however to write it as a fraction. The more correct way to write this would be $(x\cdot y^{-1})^3\equiv 2\pmod{3}$

which is impossible because of $2$ not being a cubic residue mod9. So now we know $3$ divides either $y$ or $z$ but not both.

The bit about $2$ not being a cubic residue is fine, but how do we know that $3$ divides either $y$ or $z$ but not both? All of our work so far has been under the assumption that $3$ did not divide either of $y$ or $z$, so how can any of it have any influence on the validity of the cases where $3$ does divide either of $y$ or $z$ or both. These must be treated separately.

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Suppose $3\mid z\implies 3\mid x-y$

This is the same assumption you started the first part of the proof with written in a different way.

and note: $\gcd(x-y,z^3/(x-y))=\gcd(x-y,(x^3-y^3)/(x-y))=\gcd(x-y,3x^2)=3$

You seem to be making the same mistake as before. The assumption $3\mid z\implies 3\mid x-y$ does not mean the same thing as the assumption $3\mid z$ and $3\mid x-y$. The assumption $3\mid z\implies 3\mid x-y$ is the same as the assumption $3\nmid z$ or $3\mid x-y$.

Ignoring division by zero errors because a division by zero error would imply that one of our variables was zero (which still should be pointed out during the proof, not left to the reader to notice on their own), making it to $\gcd(x-y,z^3/(x-y))=\gcd(x-y,3x^2)$ is fine. As written however, your assumption was not explicitly that $3\mid x-y$, it could still be that $3\nmid z$ and $3\nmid x-y$ in which case $\gcd(x-y,3x^2)=3$ is false. If we change your hypothesis to be $3\mid z$ and $3\mid x-y$ instead we continue:

So we know $9\nmid x-y$

What? Where did this come from? Knowing that $3$ is the greatest common divisor between $x-y$ and $3x^2$ doesn't directly imply that $9$ is not a divisor of $x-y$. It could be that $9$ is a divisor of $x-y$ but not a divisor of $3x^2$.

Take the following equation $(x-y)^3+3y(x-y)^2+3y^3(x-y)=x^3-y^3=z^3\equiv 0\pmod{27}$

$\implies 3y^2(x-y)\equiv 0\pmod{27}\implies 3\mid y$ contradicting $\gcd(x,y,z)=1$

This follows from your given work and (modified) assumptions so far, but this does not cover many of the remaining cases.