Solve $x^2+2=y^3$ using infinite descent?

Just so this doesn't get deleted, I want to make it clear that I already know how to solve this using the UFD $\mathbb{Z}[\sqrt{-2}]$, and am in search for the infinite descent proof that Fermat claimed to have found.

I've alaways been fascinated by this Diophantine equation $x^2+2=y^3$ in particular ever since I saw it, and I still have no clue how to attack it without $\mathbb{Z}[\sqrt{-2}]$. What's disappointing is that no one else seems interested in the hunt (an elementary proof using infinite descent). I know it's been studied extensively, and there have even been generalizations, such as Mordell's equation. However, I've never seen Fermat's original proof that $(x,y)=(\pm 5, 3)$ is the only integer solution. Obviously, Fermat probably knew nothing of UFD's, which is why I believe there has to be an infinite descent proof like he claimed. Has anyone apart from him actually seen this proof? People mention it all the time, yet I can't find anything about it. As I said, I know that it involved infinite descent, but I've never seen it anywhere and no one seems to have any idea about it.

Does anyone have ideas for this approach? I mean, infinite descent seems more effective for showing a contradiction, e.g. showing there are no solutions. But how could it work here? Also, why isn't it published anywhere in all this time? Could it really be that only Fermat knew his method of descent well-enough to make this problem submit to it?

Thanks!

To answer your last question: Yes, it could really be that only Fermat knew his method of descent [using contemporary techniques] well enough to make this problem submit to it.

The companion problem regarding $x^2+4=y^3$ also has no known descent proof, though he claimed to have one. There is no known descent proof of the fact that Pell's equation has infinite solutions — but Fermat claimed to have proven that by descent as well. In fact, of the ten problems mentioned in his letter to Carcavi, which Fermat claimed to prove by infinite descent, as far as I know only one (FLT for $n=3$) has had a published descent proof.

To summarize: If Fermat had only claimed to have proven one of his theorems (e.g. FLT) by descent, and no such proof was ever found, I would have no problem convincing myself that he was mistaken. But he claimed descent proofs of dozens of theorems, all of which were later proven true using other methods — at some point, we have to ask ourselves what he knew that we don't.

Even though you've already accepted my other answer (to your second/last question), I thought I’d add some thoughts on how the equation $x^2+2=y^3$ might be attacked by descent.

This answer is Method #1. I would love to brainstorm how this might be completed, or why it cannot.

Rewrite the equation as \begin{align} x^2+3 &= y^3+1 = (y+1)(y^2-y+1). \end{align} We can show that $3 \nmid x(y+1)$, and that $x,y$ are odd, and hence that $\gcd(y+1,y^2-y+1)=1$. Therefore by well-known results, we have \begin{align*} y+1 &= a^2 + 3b^2, \\ y^2-y+1 &= c^2 + 3d^2 \end{align*} for integers $a,b,c,d$ with $ac \ne 0$. (Since $y^2-y+1 = \tfrac{1}{4}\bigl((y+1)^2 +3((y+1)-2)^2\bigr)$, we can actually define $c,d$ in terms of $a,b$, but for now this should give you an adequate idea of my suggested approach.) Multiplying gives \begin{align} x^2 + 3 = (a^2+3b^2)(c^2+3d^2) = (ac \pm 3bd)^2 + 3(ad \mp bc)^2, \end{align} which can be rewritten as \begin{align} x^2 - 3(ad \mp bc)^2 = (ac \pm 3bd)^2 - 3(1)^2 = k, \tag{$\star$} \end{align} where $k \le x^2-3$ is an unknown integer. Now ($\star$) gives two solutions to the equation $U^2-3V^2 = k$, and $(ac \pm 3bd,1)$ with one of the signs is the fundamental solution (because of the $1$ at the end).

My intuition says that we can apply some sort of descent on ($\star$), and ultimate deduce that $a=\pm 2$ and $b=0$, forcing $y=a^2+3b^2-1=4+0-1=3$, as desired.

See this thread for more.

Potential Descent Mechanism #5.

For $Y^2+2=X^3$, the only solution is $(y,x)=(5,3)$. Note that $3=1+2=1^2+2$. Hence writing $z=1=\sqrt{x-2}$, we have $$y^2+2=(z^2+2)^3.$$

For $Y^2+4=X^3$, the only solution with odd $x,y$ is $(y,x)=(11,5)$. Note that $5=1+4=1^2+4$. Hence writing $z=1=\sqrt{x-4}$, we have $$y^2+4=(z^2+4)^3.$$

Once each is in this form, it’s actually quite easy to show that $z=1$ (though I’ve not found a descent proof, only several different direct proofs). Perhaps there is a way to get to these forms by infinite descent, and then perhaps from these forms to the desired conclusion $z=1$ also by infinite descent?

Here’s Potential Descent Mechanism #2 for discussion.

Clearly, $x$ and $y$ are both odd, and $x > y$. Hence there exist integers $a > b \ge 1$ such that $x=a+b$ and $y=a-b$. After substitution, you end up with a new third-degree equation, which I believe to be more susceptible to attack due to the number of terms and cross terms.

See this MSE thread, and this one, and this MO thread for examples of me trying — unsuccessfully — to apply higher-degree Vieta-jumping to obtain a descent path against such equations. I still think it's possible.

Potential Descent Mechanism #3.

You're right that infinite descent seems more effective for showing a contradiction, e.g., showing there are no solutions.

But we aren't required to prove there are no solutions to the original equation… We can instead prove that there are no solutions to some implied equation, the [hypothetical] nonzero solutions of which would be solutions to the original equation with (e.g.,) $x>5$.

As one example, write $(x,y) = (5+2u,3+2v)$, where $u,v$ are positive integers by hypothesis. Now by substituting and simplifying, we have the equation. $$2u(u+5) = v(4v^2+18v+27).$$ Here, we need to prove that $v=0$ and $u \in \{0,-5\}$, i.e., the equation has no positive integer solutions. That is likely [more] susceptible to attack by descent.