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Among the great mathematicians who struggled to try and figure out a proof of Fermat's last theorem in 19th century, Lamé was probably one of the most convinced in having succeded. I first came across this topic when reading the book $\textit{Fermat's Last Theorem. A Genetic Introduction to Algebraic Number Theory}$. After a brief search for details of Lamé's proof, I found in What was Lame's proof? the steps he followed, but, even in its fault, it does not work if $n=3$, as the author himself claimed, in $x^n+y^n=z^n$, the equation in Fermat's Last Theorem statement.

I found that the result for $n=3$ was already proven to be true at the time, thanks only to the factorization of the left member of the equation, so Lamé did not actually needed to prove it. However, in many other sources, it is stated that there is a different proof which exploits the infinite descent method, like the one Lamé employed, but no one gives details about that.

The closest I came was in some notes related to a course of algebraic number theory. They stated that, working in $\mathbb{Z}[j]$, where $ j=\frac{-1+i\sqrt{3}}{3} $ is a 3rd-root of the unity in the complex field, $x^3 + y^3 = z^3$ can be rewritten as follows: $$ \left( \frac{x}{z} \right)^3 -1 = - \left( \frac{y}{z} \right)^3. \tag{1} \label{1} $$ So, by using $$X^3-1 = (X - 1)(X^2 + X + 1) = (X - 1)(X - j)(X - j^2),$$ and multiplyng by $z^3$ in $(\ref{1})$, it ensues that: $$ (x - z)(x - jz)(x -j^2z) = -y^3. $$

Unfortunately, the notes ended by claiming that, from this identity, Lamé concluded by infinite descent using the additional fact that, if the product of three coprime integers is a cube, then each of the factor can be written as a cube. Here, the three coprime elements should be $ (x - z)$, $(x - jz)$ and $(x -j^2z)$. Probably, Lamé made a mistake in extending to $\mathbb{Z}[j]$ such a property known to be true only in the integer ring, still I can't understand how the infinite descent should develop, starting from this point. Can anyone please help me to find a way in proceeding with this argument?

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  • $\begingroup$ A modern version of Lamé's proof of the so-called "first case" of Fermat's Last Theorem (where $p$ does not divide $xyz$) is given as a series of $11$ exercises in Chapter 1 of An Invitation to Arithmetic Geometry by Dino Lorenzini. $\endgroup$ – FredH Mar 26 at 22:42
  • $\begingroup$ Actually, $\mathbb{Z}[j]$ is a UFD, so the property does hold there. $\endgroup$ – Arturo Magidin Mar 26 at 22:47
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As you well know, Fermat’s equation can immediately be reduced to the case of prime exponent $p$, say $(F_p) : x^p+y^p=z^p$, with coprime integers $x,y,z$. Two cases are usually considered, the first case where $p ∤ xyz$, and the second case where $p$ divides exactly one variable, say $z$. For $p=3$, the solution is generally attributed to Euler. The proofs of the two cases are quite different.

1). First case. Consider the eq. $(F_3)$ with variables $x,y,z \in A=\mathbf Z [√3]$, which is the ring of integers of $K=\mathbf Q(√3)$ because $3≡-1$ mod $4$. The important fact is that $K$ happens to have class number 1, i.e. $A$ is a PID. Just as previously in $\mathbf Z$, suppose in the first case that $π ∤ xyz$, with $π=√3$. Since $π^2=3$, π is a prime of $A$, with ramification index 2, i.e. $ A/πA≅\mathbf Z/3$, hence $t^3≡t$ mod π for any $t∈A$. Besides, eq. $(F_3)$ implies that $x^3+y^3≡z^3$ mod $π^a$ for any $a$. Taking $a$ large, we conclude from the binomial formula $(x+y)^3=x^3+y^3+3xy(x+y)$ that $xy(x+y) ≡xyz≡0$ mod π : contradiction. One could also factorize $x^3+y^3=(x+y)(x+jy)(x+j^2y)$ in $B=\mathbf Z[j]$ and use the fact that $B$ is a PID. This second approach (Kummer) works in $\mathbf Z [\zeta_p]$ as long as $p$ is a regular prime, i.e. does not divide the class number.

2). Second case. This is notoriously more complicated. Let us first recap Euler’s original approach to $(F_3)$, with $z≡0$ mod 3. Euler introduces the auxiliary equation $(E_3) : x^2+(3y)^2=z^3$, which he solves by working in $\mathbf Z[i√3]$, asserting without proof that this ring is a PID. This is incorrect, but fortunately the ring of integers of $\mathbf Q(i√3)=\mathbf Q(j)$, which is $B= \mathbf Z[(1+i√3)/2]=\mathbf Z [j]$, is a PID. Working in $B$, it can be shown that all solutions of $(E_3)$ are of the form $x+iy√3=(a+ib√3)^3$. Putting $u=(x+y)/2$ and $v=(x-y)/2 , (F_3)$ becomes $2u(u^2+3v^2 )=z^3$. Using $u≡0$ mod 3, Euler can perform descent on⎹ xyz⎹ and conclude. Note that Kummer's proof of the second case for regular $p$ is much more elaborate.

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  • $\begingroup$ I'd like to give more upvotes: much compact and well collected information! (+1) $\endgroup$ – Gottfried Helms Mar 28 at 19:22

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