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

Question

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!

Answer 1

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.

Answer 2

A definitive (?) answer to your question can be found on page 561 of the Nov 2012 edition of The Mathematical Gazette, where a totally elementary descent mechanism first used by Stan Dolan in the March 2012 edition is adapted (as per his challenge to the “interested reader”) to solve Fermat’s two “elliptic curve” theorems. The method uses math which was clearly available in Fermat’s time, and in particular to Fermat himself.

I personally believe this finally puts to rest any questions of whether Fermat could have had a proof of these two claims.

Answer 3

Lemma. Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $n=t^2+2u^2$ divides $bt-au$.

Proof. Assume the theorem is false, and let $m$ be a minimal counterexample. Evidently $m > 1$ since the theorem is trivially true for $m=1$.

Note that $b$ is coprime to $m$. Let $A$ be an integer such that $Ab \equiv a\!\pmod{m}$, chosen so that $\tfrac{-m}{2} < A \le \tfrac{m}{2}$. Then $A^2+2 = lm$ for some positive integer $l < m$. Clearly $l$ cannot be a smaller counterexample than $m$, and so there exist coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$.

Let $t = \tfrac{ar+2bs}{m}$ and $u=\tfrac{br-as}{m}$. Direct calculation confirms the equations for $a$, $b$, and $n$. From $n=t^2+2u^2$, we deduce that $t$ is an integer because $u$ is an integer, and $t$ and $u$ are coprime because $\gcd(t,u)$ divides both $a$ and $b$. Finally, note that $n$ divides $bt-au=sn$.

Hence $m$ is not a counterexample, contradicting the original assumption. $\blacksquare$

Corollary. Let $a$ and $b$ be coprime integers with $m$ an integer such that $m^3=a^2+2b^2$. Then there are coprime integers $r$ and $s$ such that $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$.

Proof. Evidently $m$ is odd since $a^2+2b^2$ is at most singly even. And $a$ and $m$ must be coprime. Using the theorem, we have $m=r^2+2s^2$ and $m^2=t^2+2u^2$. Then $m$ divides $a(ur-ts)=t(br-as)-r(bt-au)$, and therefore $m \mid (ur-ts)$. The lemma can then be reapplied with $a$ and $b$ replaced by $t$ and $u$. Repeating the process, we eventually obtain integers $p$ and $q$ such that $p^2+2q^2=1$. The only solution is $q=0$ and $p=\pm1$. Ascending the path back to $a$ and $b$ (reversing signs along the way, if necessary) yields $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$, as claimed. $\blacksquare$

Theorem. The Diophantine equation $X^3 = Y^2+2$ has only one integer solution, namely $(x,y) = (3, \pm 5)$.

Proof. Evidently $y$ and $2$ are coprime. By the corollary, we must have $b=1=s(3r^2-2s^2)$ for integers $r$ and $s$. The only solutions are $(r,s)=(\pm 1,1)$. Hence $a=y=r(r^2-6s^2)=\pm 5$, so $(x,y)=(3,\pm 5)$. $\blacksquare$

Answer 4

Potential Descent Mechanism #4.

Observe, regarding Fermat’s “Carcavi problems”:

1. The only [nontrivial] positive solution to the equation $Y^2+2=X^3$ is $(y,x)=(5,3)$. Both elements are primes.
2. The only [nontrivial] positive integer solutions to the equation $Y^2+4=X^3$ are $(y,x)=(2,2)$ and $(y,x)=(11,5)$. In both cases, both elements are primes.
3. The only nontrivial positive integer solution to the system of equations $X+1=2Y^2$ and $X^2+1=2Z^2$ is $x=7$, which is a prime.
4. If $x^n+y^n=z^n$ with relatively prime integers $0<x<y<z$ and an integer $n \ge 3$, then neither $y$ nor $z$ can be a prime power (cf. Abel’s Conjecture, this part of which has been proven).
5. Fermat’s descent proof for his two-squares theorem descends to the known solution $5=1^2+2^2$. And (wait for it!) $5$ is a prime.

So… what if Fermat’s method of infinite descent was based not on assuming a solution and then finding a smaller solution of the same form… but rather on assuming a solution in which one of the [integer] components/elements had multiple distinct prime factors, and then showing that there must be a solution in which that same component/element was comprised of fewer primes? This would (by infinite descent) lead to the case where there must be a solution in which that component/element was a prime (or, at worst, a prime power) — and it might at that point be “small enough to drown in a bathtub”.

q.v. my question in this direction

In the particular case of your question: it is quite easy to show that $3 \mid x$ in any solution to the equation $Y^2+2=X^3$. If one could then show that $x$ must be a prime, we’re done; if $x$ must be a prime power, there’s probably a quick coup de grâce there as well (though I haven’t taken the time to find it).

Answer 5

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?

Answer 6

I think I have a plausible explanation. You're probably familiar with the generator for Pythagorean triplets:

$x^2-y^2=a \quad 2xy=b \quad x^2+y^2=c \quad \longleftrightarrow \quad a^2+b^2=c^2$

One could say this is a complete solution to the equation $a^2+b^2=c^2$ in the sense that 1) $x$ and $y$ can be chosen randomly to calculate $a,b,c$, and 2) for every example of $a^2+b^2=c^2$ where $a$ is coprime to $b$, there will always be $x$, $y$ such that one of $x^2-y^2$ or $2xy$ is $a$ and the other is $b$.

Along a similar line, complete solutions to $a^2+2b^2=m^i$, for positive $i$, can be obtained by generating two sequences of polynomials $f_1,f_2,f_3$... and $g_1,g_2,g_3$... by

$f_1 := x$
$g_1 := y$
$f_{i+1} := x(f_i)-2y(g_i)$
$g_{i+1} := x(g_i)+y(f_i)$

In other words, if you find positive coprime $a$ and $b$, and $a^2+2b^2=m^i$, there will always be $x$ and $y$ such that $x^2+2y^2=m$, $\left\vert f_i \right\vert = a$, and $\left\vert g_i \right\vert = b$.

I can prove this, and would suppose it is already well known. I generally find after any discovery that some mathematician already treated the subject more comprehensively and elegantly a few centuries ago. In any case, my proof starts with a few things proven directly, but the final part of it is by infinite descent. [3/31/2017 - Since my earlier post, I have noticed that a substantial part of what leads to the final infinite descent argument can also be accomplished with infinite descent. I would say I can rigorously prove that 27 is the only cube two greater than a square using "ninety percent" infinite descent arguments, and "ten percent" direct algebraic and logic arguments.] By that time, I have established that if there is an example of $a^2+2b^2=m^i$, with no corresponding $x$ and $y$, then $m$ must be a composite number. Furthermore, if $m$ is divisible by the prime number $p$, there will be another counter-example for $m_2 := {m \over p}$. Then there is an infinite sequence $m, m_2, m_3$, ... and every time $m_i$ is divided by a prime number, the result is $m_{i+1}$ which is composite.

I can provide further details if need be. It seems to me at least possible that Fermat used this approach. Certainly it leads with ease to the solution to $c^2+2=m^3$. In this case $a=c, b=1, i=3$. Since 1 is coprime to all numbers, $a$ is coprime to $b$. Then $f_3$ and $g_3$ should generate the example.

$$f_3 = x^3-6xy^2 = \pm c$$ $$g_3 = 3x^2y-2y^3 = \pm 1$$ $$x^2+2y^2=m$$

In the middle equation, 1 is a multiple of $y$, and the only possibilities for $y$ are 1 or -1. That leads to the only solutions being $x=\pm1, y=\pm1$, and all four of these lead to the same end result, $c=5, m=3$.

The equation $c^2+4=m^3$ can be treated in a similar manner, by extending the Pythagorean triplet generator. In this case use $f_{i+1} := x(f_i)-y(g_i)$ instead of what is given above, but keep the definition for the $g$'s. However, there is a slight additional complication. Solving $g_3=\pm 2$ will lead to the example $11^2+4=5^3$ as well as the example $2^2+4=2^3$ in which $a$ is not coprime to $b$. However, 2 is not coprime to all numbers, and you will need to additionally demonstrate that 2 is the only even value for $c$ such that $c^2+4$ is a cube. Note the stipulation that $a$ is coprime to $b$ is necessary. For example, $9^2+12^2=15^2$ has no $x,y$ because 15 is not the sum of two squares.

Answer 7

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.

Answer 8

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.

Answer 9

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.