Question on a crucial lemma in Euler's approach to Fermat's Last Theorem for $n=3$

Question

As many of you may know, the illustrious L. Euler put forward a proof of the case $n=3$ of Fermat's Last Theorem via infinite descent. The thing is that, at a certain point, he resorted to the following result:

LEMMA. Let $a$ and $b$ be two coprime integers such that $a^2+3b^2=z^{3}$ for some $z\in \mathbb{Z}$. Then, it is possible to find integers $s$ and $t$ such that

$$a= s(s^2-9t^2) \qquad \mbox{and} \qquad b=3t(s^2-t^2).$$

If I understand correctly, Euler tried to settle this lemma by factoring $a^2+3b^2$ as $(a+b\sqrt{-3})(a-b\sqrt{-3})$, writing $z=u+v\sqrt{-3}$ for some integers $u$ and $v$, and pretending afterwards that a third powers product principle holds in $\mathbb{Z}[\sqrt{-3}]$. Just in case, this is what I have in mind when I write of a third powers product principle in a ring $R$ (I suppose the name is more or less well-established; I picked it up from P. Pollack's "A conversational intro. to algebraic number theory"):

If $\alpha$ and $\beta$ are nonzero elements of $R$ whose product is the cube of an element $\gamma$ of $R$ and $\alpha$ and $\beta$ have no common divisors other than units, then both $\alpha$ and $\beta$ are cubes of elements of $R$ (up to multiplication by a unit).

Unfortunately, it was eventually recognized that Euler's attack on the lemma was flawed precisely because of the (implicit) assumption regarding the veracity of a third powers product principle in $\mathbb{Z}[\sqrt{-3}]$.

Having said all this, I wonder if you know of a reference wherein one can find a transparent proof of the lemma... To be a hundred percent honest, I find a wee bit cumbersome the one that we can read in section 2.5 of H. M. Edwards book on Fermat's Last Theorem.

Please, let me thank you in advance for your insightful comments and all the bibliographical information regarding this lemma that you may want to share with me.

Answer 1

Euler's attack on the lemma was flawed precisely because of the (implicit) assumption regarding the veracity of a third powers product principle in $\mathbf Z[\sqrt{-3}]$.

A close reading of Euler's proof shows that he needs this property for numbers in $\mathbf Z[\sqrt{-3}]$ with odd norm, and for such numbers that property in fact is true. That is, numbers in $\mathbf Z[\sqrt{-3}]$ with odd norm have unique factorization. Proving that makes Euler's proof correct, but it is simpler instead to work in the slightly larger ring $\mathbf Z[\zeta_3]$ that is a UFD and use knowledge of its units to pass back to $\mathbf Z[\sqrt{-3}]$.

Answer 2

You can simply work in the unique factorization domain $R:=\mathbb{Z}[\omega]$, where $\omega=(-1+\sqrt{-3})/2$ is a cubic root of 1. $R$ has 6 units $\pm 1$, $\pm \omega$, $\pm \omega^2$.

If 3 divides $a$, then 3 does not divide $b$ and $z^3$ is divisible by 3 but not by 9, a contradiction. If $a,b$ are both odd, then $z^3$ is congruent to 4 modulo 8, a contradiction. So, one of $a$, $b$ is odd and another is even, and $z$ is odd.

Now denote by $d$ the common divisor of $u:=a+b\sqrt{-3}=a+b(2\omega+1)$ and $v:=a+b\sqrt{-3}=a-b(2\omega+1)$ in the ring $R$. Then $d$ divides $2a=u+v$ and also $d$ divides $uv=a^2+3b^2=z^3$. Since $z$ is odd, $d$ is coprime with 2, thus $d$ divides $a$. But $a$ is coprime both with 3 and with $b$, thus $a$ is coprime with $3b^2$ and therefore $a$ is coprime with $a^2+3b^2$. Hence $d=1$. Therefore $u=\varepsilon (x+y\omega)^3$, where $x,y$ are integers and $\varepsilon$ is a unit. Note that $u\equiv 1\pmod 2$. We have $(x+y\omega)^3=x^3+y^3+3x^2y\omega+3xy^2\omega^2=(x^3+y^3-3xy^2)+3\omega(x^2y-xy^2)$. Since $x^2y-xy^2$ is always even, we see that $(x+y\omega)^3$ is congruent to 0 or 1 modulo 2. And being multiplied by $\varepsilon$ this must be congruent to 1 modulo 2. This is only possible if $\varepsilon=\pm 1$. Since $-1$ is a cube itself, we may suppose that $\varepsilon=1$. One of three numbers $x+y\omega$, $\omega(x+y\omega)=-y+\omega(x-y)$, $\omega^2(x+y\omega)=y-x-x\omega$ (they have the same cube equal to $u$) has an even coefficient of $\omega$. Without loss of generality, this is $x+y\omega$: $y$ is even. Then $x+y\omega\in \mathbb{Z}[\sqrt{-3}]$, and denoting $x+y\omega=s+t\sqrt{-3}$ we get desired representation.

Answer 3

As far as I know, Euler considered any coefficients. And he had to solve various equations of the form.

$X^2+qY^2=Z^n$ Or as in this case, the equation of the 3rd degree. $X^2+qY^2=Z^3$

And Euler decided to find out and deal with the simplest equation.

$$X^2+qY^2=Z^3$$

To do this, I decided to use the Identity of the Brahmaputra.

$$q(ac+nbd)^2+n(ad-qbc)^2=(a^2+nqb^2)(qc^2+nd^2)$$

Or as in this case.

$$(ac+qbd)^2+q(ad-bc)^2=(a^2+qb^2)(c^2+qd^2)$$

Further transformations were based on the idea that it was necessary to solve the Pythagorean triple. $Z=a^2+qb^2$ $$c^2+qd^2=Z^2=(a^2+qb^2)^2$$

This idea made it possible to solve any equation of the form... $X^2+qY^2=Z^n$

We need to understand Euler's ideas. He believed that the simpler the transformations, the better it was. Only later it turned out that this is not always good. Especially when it can be solved in another way and there will be more parameters defining the solution.

To describe the solutions of the equation. $$X^2+qY^2=Z^3$$

I think best would be to describe a solution using $3$ parameters.

$$X=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^2)^3$$

$$Y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$

$$Z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$

You can write this simple solution:

$$X=(p^2+qs^2)((p^4-q^2s^4)t^3-3(p^2+qs^2)^2kt^2+3(p^4-q^2s^4)tk^2-(p^4-6qp^2s^2+q^2s^4)k^3)$$

$$Y=2ps(p^2+qs^2)((p^2+qs^2)t^3-3(p^2+qs^2)tk^2+2(p^2-qs^2)k^3)$$

$$Z=(p^2+qs^2)((p^2+qs^2)t^2-2(p^2-qs^2)tk+(p^2+qs^2)k^2)$$

You can try to find some beautiful transformation. Euler used the simplest. And you can choose the shape we need. For example, this.

$$x^2+3y^2=ab$$

Then the solution can be represented as:

$$x=(p-s)kn+(p^2+5ps)n^2$$

$$y=pkn+(p^2-2ps+2s^2)n^2$$

$$a=k^2+2pkn+(p^2+12s^2)n^2$$

$$b=((2p-s)n)^2$$

Or so:

$$x=k^2-2(4p+s)kn+(7p^2+26ps-8s^2)n^2$$

$$y=k^2-4(2p-s)kn+(7p^2-16ps+4s^2)n^2$$

$$a=4(k^2-(2p-s)kn+(p^2-ps+7s^2)n^2)$$

$$b=(k-(7p-2s)n)^2$$

Then it turned out that the representation of numbers. Rather, the search for such parameterizations is a separate topic.

$$(x^2+ay^2)(u^2+bv^2)=p^2+cq^2$$

https://artofproblemsolving.com/community/c3046h1053530_another_solution_almost_pythagoras_3

https://artofproblemsolving.com/community/c3046h1053462_another_solution_almost_pythagoras

https://artofproblemsolving.com/community/c3046h1053193_almost_pythagoras

But the question is not limited to this. Applying a similar approach to solving other problems, it became necessary to solve systems of nonlinear equations. But this is a much more difficult task.