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In the mid-1600s, Fermat made the following claim:

Theorem: The Diophantine equation $$Y^2+2=X^3 \tag{$\star$}$$ has only one positive integer equation, namely $(x,y)=(3,5)$.

Proofs were found by Euler and others — starting about a century after Fermat made his claim — but none of them used math that was available to Fermat. Because ($\star$) is an example of an elliptic curve, many mathematicians didn’t think Fermat had a proof, but a completely elementary proof within Fermat’s grasp was recently found (bonus: it uses uses infinite descent!).

Regardless, I’m still interested in finding alternative elementary techniques that would have been available to Fermat and his contemporaries, which might be applied to equations like ($\star$) — for example, see this question, where I attempt (in vain) to solve it another way.

I‘ve found a slight adaptation of that proof which appears [to me] to be valid, and I’d like to see if anyone can find a flaw.

Proof. Considering ($\star$) modulo $4$, it’s clear that $x$ and $y$ are both odd; furthermore, evidently $y>x$. Hence there exist integers $a>b\ge 1$ such that $x=a-b$ and $y=a+b$. Substituting into ($\star$) and simplifying yields

$$b^3-3ab^2+b^2+3a^2b+2ab-a^3+a^2+2=0.$$

This can be rewritten as \begin{align} 2(a-b)(a^2+b^2+1) &= (a-b+1)((a+b)^2+2), \end{align} and since $a-b>0$, this implies \begin{align} \tag{$\dagger$} \frac{2(a^2+b^2+1)}{(a+b)^2+2} &= \frac{a-b+1}{a-b}. \end{align} A quick calculation on ($\star$) shows that in the range $1 \le a-b \le 3$, the only solution is $a-b=3$, which implies $x=3$. Now assume $a-b=x > 3$. As $a-b$ approaches infinity from below, the right-hand side of ($\dagger$) approaches $1$ from above; hence the left-hand side must also approach $1$ from above. Using $\rightarrow$ to mean “approaches from above”, we have the following: \begin{align} \frac{2(a^2+b^2+1)}{(a+b)^2+2} &\rightarrow 1 \\[0.5em] 2(a^2+b^2+1) &\rightarrow (a+b)^2+2 \\ a^2+b^2 &\rightarrow 2ab \\ (a-b)^2 &\rightarrow 0. \end{align} Since $a-b > 3$ and growing, this is clearly impossible. Hence the only solution is $x=3$. QED

Where is the error in this logic?

Kieren MacMillan
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1 Answers1

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There are two serious errors in your reasoning.

The first is that $a,b$ are fixed, so you can't assume $a-b$ approaches infinity.

Ignoring that error, another serious error is your work on the LHS.

A limit statement of the form $$ \frac{u}{v}\rightarrow 1 $$ does not imply $ u-v\rightarrow 0 $.

Kieren MacMillan
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quasi
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  • Worrying just about the first issue for this comment… Since $x=a-b$, is there a way to avoid the problem? For example, would it be rigorous to say something like “For any solution $x = a-b > 3$, the RHS is closer to 1 than in the case $a-b=3$, and hence the left-hand side implies [show math here] that $a-b$ is closer to 0”? – Kieren MacMillan Aug 10 '21 at 17:18
  • The second issue sure seems like a showstopper… I’ll think more about that. Thanks – Kieren MacMillan Aug 10 '21 at 17:18
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    @Kieren MacMillan: Regarding the first issue, you could try to show that for $a-b$ greater than some specified threshold, the LHS is always, say, greater than the RHS. But that idea is doomed to failure since the original equation has real solutions for all $x > 2$. There's no contradiction if all you assume is that your variables are real. – quasi Aug 10 '21 at 17:37