Fermat's last theorem (Not to freak, just a special one)

[mathbalarka]

The NT is a little down on MHB lately. Let's try to do some stuffs to bring the class a wee bit back : prove that there are no nontrivial integer solutions to

$$x^3 + y^3 = z^3$$

Well, this is a decent and well-known diophantine form and there are many ways to prove it, not to mention assuming the more general FLT itself (but let's step aside from that, as it's too complicated and the margin on the form is too short)

Flood in, folks!

Edit : Okay, we need some rule since this case is a well-known one. How about "Only one solution from each participant". Have fun inventing new methods. Now you have to choose the methods you find fits best. I am evil, aren't I? (Devil)

 

[mente oscura]

Re: Fermats last theorem (Not to freak, just a special one)

The NT is a little down on MHB lately. Let's try to do some stuffs to bring the class a wee bit back : prove that there are no nontrivial integer solutions to

$$x^3 + y^3 = z^3$$

Well, this is a decent and well-known diophantine form and there are many ways to prove it, not to mention assuming the more general FLT itself (but let's step aside from that, as it's too complicated and the margin on the form is too short)

Flood in, folks!

Edit : Okay, we need some rule since this case is a well-known one. How about "Only one solution from each participant". Have fun inventing new methods. Now you have to choose the methods you find fits best. I am evil, aren't I? (Devil)

Hello. (Movie) (Rofl)

Spoiler

$$Let \ a, \ b, \ c, \ g, \ r \in{N} \ / \ a, b=odd \ c,t=even \ / a,b,c \ coprime$$

$$Supposing \ a^3=b^3+c^3$$

$$\dfrac{a^3}{b+c}=\dfrac{b^3+c^3}{b+c}=b^2-bc+c^2=(b+c)^2-3bc$$

We can deduce that:

$$b+c=g^3 \ and \ (b^2-bc+c^2)=r^3$$

$$Let \ x \in{Z}/ r=b+x$$

$$(b+x)^3=b^2-bc+c^2 \rightarrow{} (b+x)|(b^2-bc+c^2)$$

Dividing:

$$\dfrac{b^2-bc+c^2}{b+x}=b-(c+x)+\dfrac{[c^2+x(c+x)]}{b+x}$$

$$Rest=0 \rightarrow{} c^2+x(c+x)=c^2+cx+x^2=0$$

$$x=\dfrac{-c \pm \sqrt{c^2-4c^2}}{2}\rightarrow{}x \cancel{\in}R$$

I ask forgiveness, for using an own nomenclature.

Regards.

Edit:

Spoiler

possible exception:

$$3|(b+c)$$

But, i could try with:

$$\dfrac{b^3}{a-c}=\dfrac{a^3-c^3}{a-c}=a^2+ac+c^2$$

Etc...

 

[mathbalarka]

Re: Fermats last theorem (Not to freak, just a special one)

It's not necessarily true that $b + c$ and $b^2 - bc + c^2$ are coprime, in which case the'd be cubes. I'd have done it a little differently. Hint : Split the second factor in $\Bbb Z[\zeta_3]$

 

[mente oscura]

Re: Fermats last theorem (Not to freak, just a special one)

It's not necessarily true that $b + c$ and $b^2 - bc + c^2$ are coprime, in which case the'd be cubes. I'd have done it a little differently. Hint : Split the second factor in $\Bbb Z[\zeta_3]$

They are necessarily coprime, with the exception that they are divisible by "3". Then the study would be done with the part that I have edited.

Spoiler

$$b^3+c^3=(b+c)[(b+c)^2-3bc]$$

$$Let \ m \in{N} \ / \ m|(b+c) \ and \ m \cancel{|}(3bc)$$. Exception: $$3|(b+c)$$

In that case, i would demonstrate it by the same procedure, with (a-c), as i express in the edited part.

 

[mathbalarka]

Re: Fermats last theorem (Not to freak, just a special one)

They are necessarily coprime

Prove it, then.

with the exception that they are divisible by "3"

Then they are not coprime :D

 

[mente oscura]

Re: Fermats last theorem (Not to freak, just a special one)

Spoiler

If $$(b+c) \ and \ (b^2-bc+c^2) \ not \ coprime \ \rightarrow{}3|(b+c) \ and \ 3|)b^2-bc+c^2$$

Then $$(a-c) \ and \ (a^2+ac+c^2) \ coprime$$

$$\dfrac{b^3}{a-c}=\dfrac{a^3-c^3}{a-c}=a^2+ac+c^2$$

$$a-c=f^3 \ and \ (a^2+ac+c^2)=q^3$$

$$Let \ y \in{Z}/ (a-y)^3=q^3=a^2+ac+c^2$$

$$(a-y)|(a^2+ac+c^2)$$

Dividing:

$$\dfrac{a^2+ac+c^2}{a-y}=a+(c+y)+\dfrac{y(c+y)+c^2}{a-y}$$

$$Rest=0 \rightarrow{}y(c+y)+c^2=0$$

$$y=\dfrac{-c \pm \sqrt{c^2-4c^2}}{2} \cancel{\in}R$$

Testing:

1º) $$3|a \ or \ 3|b \ or \ 3|c$$

Demonstration:

$$Let \ t \in{N} \ / \ a+t=b+c$$

$$t^3=(b+c-a)^3$$

calculating:

$$t^3=3(a-b)(a-c)(b+c)$$

Therefore:

$$3|(a-b) \ or \ 3| (a-c) \ or \ 3|(b+c)$$

Therefore:

$$3|a \ or \ 3|b \ or \ 3|c$$

2º)

$$Let \ E, \ F, \ G, \ P, \ Q, \ R \in{N} \ /$$

$$/ \ E=(a-b), \ F=(a-c), \ G=(b+c), \ and$$

$$and \ P=\dfrac{c^3}{a-b}, \ Q=\dfrac{b^3}{a-c}, \ R=\dfrac{a^3}{b+c}$$2.1)

$$\dfrac{a^3}{b+c}=\dfrac{b^3+c^3}{b+c}=b^2-bc+c^2=(b+c)^2-3bc=R$$

$$(b+c) \ and \ R \ coprime \rightarrow{}G=g^3 \ and \ R=r^3$$

Exception: $$3|(b+c)$$

2.2)

$$\dfrac{b^3}{a-c}=\dfrac{a^3-c^3}{a-c}=a^2+ac+c^2=Q$$

$$(a-c) \ and \ Q \ coprime \rightarrow{}F=f^3 \ and \ Q=q^3$$

Exception: $$3|(a-c)$$2.3)

$$\dfrac{c^3}{a-b}=\dfrac{a^3-b^3}{a-b}=a^2+ab+b^2=P$$

$$(a-b) \ and \ P \ coprime \rightarrow{}E=e^3 \ and \ P=p^3$$

Exception: $$3|(a-b)$$But the exception can only be in one of the three: 2.1), 2.2) or 2.3), (a, b, c, coprime).

 

[Deveno]

Re: Fermats last theorem (Not to freak, just a special one)

I'd like to address this problem, by way of searching for a minimal counter-example. This may span several posts, as I may need more time than I have at the moment to complete it. (Note: the argument itself is not mine, but one I have seen and worked through. I found it challenging to understand, so I hope I do it justice).

[sp]Before I begin, I will present certain facts without proof:

Consider the complex number:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$.

This number has some nice algebraic properties:

1. $\omega^3 = 1$
2. $\omega^2 + \omega = -1$

We will be working in the ring $\Bbb Z[\omega]$ which consists of all complex numbers of the form:

$a + b\omega,\ a,b \in \Bbb Z$.

We will actually prove that there is no solution to:

$x^3 + y^3 = z^3$ for $x,y,z \in \Bbb Z[\omega]$ which will then imply there are no integer solutions, since:

$\Bbb Z \subset \Bbb Z[\omega]$.

We observe that the complex norm $d(z) = \|z\|^2$ induces a norm on our ring, which is multiplicative by dint of the same property for the complex numbers. We will use the fact that this norm makes $\Bbb Z[\omega]$ into a Euclidean domain, and thus is a fortiori a unique factorization domain.

Explicitly: $d(a + b\omega) = a^2 - ab + b^2$ as one can verify by putting this number in complex form in the basis $\{1,i\}$. Note that this is always a non-negative integer.

The most important tool we will use is that, in $\Bbb Z[\omega]$:

$x^2 - xy + y^2 = (x\omega + y\omega^2)(x\omega^2 + y\omega)$

which will prove to be very useful to us.

There is another number we will make use of:

$\lambda = 1 - \omega$.

Note that $\omega = 1 -\lambda$ as well.

Our first result:

Lemma 1: $\lambda$ is prime in $\Bbb Z[\omega]$ and $\lambda|3$.

Proof: Suppose for $a+b\omega,c+d\omega \in \Bbb Z[\omega]$ we have:

$(a+b\omega)(c+d\omega) = \lambda$

Then $3 = d(\lambda) = d(a+b\omega)d(c+d\omega)$.

As 3 is a prime integer, we have wlog, $d(a+b\omega) = 1$. This implies that either $a = \pm 1, b = 0$ or $b = \pm 1, a = 0$, that is, $a + b\omega$ is a unit. Thus $\lambda$ is irreducible in $\Bbb Z[\omega]$, and since this is a UFD, it is prime.

Finally, $\lambda^2 = (1 - \omega)^2 = 1 - 2\omega + \omega^2$

$= 1 - 2\omega -\omega - 1 = -3\omega$, so

$3 = \lambda(-\lambda\omega^2)$ showing $\lambda|3$ (this also shows that $\lambda^2|3$ as well).

Lemma 2: For any element $\mu \in\Bbb Z[\omega]$, either:

$\mu \equiv 0 \text{ (mod }\lambda)$
$\mu \equiv -1 \text{ (mod }\lambda)$ or
$\mu \equiv 1 \text{ (mod }\lambda)$

Proof: $a+b\omega = a+b(1-\lambda) = a+b - b\lambda$ that is:

$a+b\omega \equiv a+b \text{ (mod }\lambda)$

Since $\lambda|3$ and $a+b \equiv -1,0,$ or $1\text{ (mod }3)$, we are done.

Corollary: $\lambda|(\alpha - 1)(\alpha)(\alpha+1)$ for any $\alpha \in \Bbb Z[\omega]$.

Theorem 1: If $\xi \in \Bbb Z[\omega]$ is such that $\lambda \not\mid \xi$, then:

$\xi^3 \equiv \pm 1\text{ (mod }\lambda^4)$

Proof: By Lemma 2, we have: $\xi \equiv \pm 1\text{ (mod }\lambda)$ I will show the proof for $\xi \equiv 1\text{ (mod }\lambda)$, the proof for $\xi \equiv -1\text{ (mod }\lambda)$ is similar and left to the reader (some sign changes must be made).

Since $\xi = 1 + \alpha\lambda$, we have:

$\xi^3 - 1 = (\xi - 1)(\xi - \omega)(\xi -\omega^2)$

$= (1 + \alpha\lambda - 1)(1 + \alpha\lambda - (1 - \lambda))(1 + \alpha\lambda - \omega^2)$

$= (\alpha\lambda)((\alpha + 1)\lambda)(1 + \alpha\lambda +\omega +1)$

$= (\alpha\lambda)((\alpha + 1)\lambda)(1 + \alpha\lambda + 1 - \lambda + 1)$

$= (\alpha\lambda)((\alpha + 1)\lambda)((\alpha - 1)\lambda + 3)$

$= \lambda^3(\alpha)(\alpha + 1)(\alpha - 1 + \lambda\beta)$

for some $\beta \in \Bbb Z[\omega]$ since $\lambda^2|3$ (see above).

Now $\alpha - 1 + \lambda\beta \equiv \alpha - 1\text{ (mod }\lambda)$

so we see that $\lambda^4|(\xi^3 - 1)$, that is: $\xi^3 \equiv 1\text{ (mod }\lambda^4)$ QED.

(see Lemma 2 and the Corollary if this is not clear).

We'll conclude *this* post with something that looks like we're making progress:

Corollary:

If $x,y,z \in \Bbb Z[\omega]$ are such that:

$x^3 + y^3 + z^3 = 0$, then $\lambda$ divides one of $x,y,z$.

Proof: Assume that the contrary is so. Then we have, in light of Theorem 1:

$x^3,y^3,z^3 = \pm 1\text{ (mod }\lambda^4)$.

There are only four possibilities:

$x^3 + y^3 + z^3 \equiv -3\text{ (mod }\lambda^4)$
$x^3 + y^3 + z^3 \equiv -1\text{ (mod }\lambda^4)$
$x^3 + y^3 + z^3 \equiv 1\text{ (mod }\lambda^4)$
$x^3 + y^3 + z^3 \equiv 3\text{ (mod }\lambda^4)$

Only the first and fourth cases give us a sum congruent to 0 (mod $\lambda$), which we must have since $\lambda|0$. So we are left with showing that:

$\lambda^4 \not\mid 3$. As $d(\lambda^4) = 81$ while $d(3) = 9$ this is clearly impossible.[/sp]

I'll rest a bit, and let this digest.

 

[Deveno]

Re: Fermats last theorem (Not to freak, just a special one)

OK, so where was I?

[sp] We have shown that we have a certain prime $\lambda \in \Bbb Z[\omega]$ guaranteed to divide one of $x,y,z$ when:

$x^3 + y^3 = z^3$.

Without loss of generality, we will say $\lambda|z$.

(If, for example, $\lambda|x$ instead, we could consider the equation:

$z^3 + (-y)^3 = x^3$ which is clearly an equation of the same form).

A short digression to reflect on where we are headed:

We want a certain "minimal counter-example" but we haven't yet said what that MEANS. Clearly, for any such non-trivial (see discussion below) solution $x,y,z$ where:

$x^3 + y^3 = z^3$

there is some power $k$ of $\lambda$ such that:

$\lambda^k|z$ but $\lambda^{k+1}\not\mid z$

(again, this is because of unique factorization into primes).

By a minimal solution, we mean one where $k$ is minimal. The set of all such $k$ is some subset of the natural numbers, and it is bounded below by 1. It is a consequence of the well-ordering of the natural numbers then that the set:

$\{k \in \Bbb Z^+: \exists x,y,z \in \Bbb Z[\omega]^{\ast}\text{ with } x^3 + y^3 = z^3 \text{ and } \lambda^k|z,\lambda^{k+1}\not\mid z\}$

has a least element provided this set is not empty (which is what we are assuming, hoping to derive a contradiction).

One more caveat: by "solution" we mean "non-trivial" solution, that is, we are disallowing one (or more) of $x,y,z$ to be 0 (for if $z = 0$, for example, EVERY power of $\lambda$ divides it, and this is not particularly insightful...we are not interested in the solutions:

$x^3 + (-x)^3 = 0^3$ or:
$x^3 + 0^3 = x^3$, or:
$0^3 + 0^3 = 0^3$.

Another caveat, we will require that $,x,y,z$ be pairwise co-prime. It is easy to see that if $\pi$ is a prime of $\Bbb Z[\omega]$ and $\pi$ divides any two of $x,y,z$ it divides all three, so we can divide (perhaps repeatedly) through by $\pi$ to obtain another solution where this does not occur. So, just to be clear, henceforth when we say "solution to Fermat's Last Theorem (for $n = 3$)" we mean "non-trivial solution where $x,y,z$ are pairwise co-prime". So let's continue.

Since $\Bbb Z[\omega]$ is Euclidean, it is a GCD domain, so we may invoke Bezout's Identity:

$\text{gcd}(\alpha,\beta) = 1 \iff \exists u,v \in \Bbb Z[\omega]: u\alpha + v\beta = 1$.

In particular, for our assumed minimal solution, $x,y,z$, we have:

$ux + vy = 1$ for some $u,v \in \Bbb Z[\omega]$.

Now:

$z^3 = x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)(x\omega + y\omega^2)(x\omega^2 + y\omega)$

as we indicated in our last post.

Since $\lambda|z$, and $\lambda$ is prime, $\lambda$ divides the product on the right.

Lemma 3: $\lambda = \text{gcd}((x+y),(x\omega + y\omega^2),(x\omega^2 + y\omega)$

Proof: Let $u,v \in \Bbb Z[\omega]$ be such that: $ux + vy = 1$. Then

$\lambda = (ux + vy)(\lambda) = (ux + vy)(1 - \omega)$

$= ux + vy - ux\omega - vy\omega$

$=vx +vy - ux\omega - uy\omega + ux\omega^3 + uy\omega^4 - vx\omega^3 - vy\omega^4$ (the 4 extra terms cancel)

$= (v - u\omega)(x + y) + \omega^2(u - v)x\omega + y\omega^2)$.

Hence any common divisor of $(x+y),(x\omega + y\omega^2)$ divides $\lambda$.

Similarly:

$\lambda = (v\omega - u)(x\omega + y\omega^2) + (u\omega - v)(x\omega^2 + y\omega)$

$\lambda = \omega^2(v - u)(x\omega^2 + y\omega) + (u - v\omega)(x + y)$

So we see any common divisor of $(x+y),(x\omega + y\omega^2),(x\omega^2 + y\omega)$ divides $\lambda$, which establishes our claim.

Again, since $\Bbb Z[\omega]$ is a UFD, this means we can find $\alpha,\beta,\gamma \in \Bbb Z[\omega]$ such that:

$x + y = \lambda\alpha,\ x\omega + y\omega^2 = \lambda\beta, x\omega^2 + y\omega = \lambda\gamma$

and $\alpha,\beta,\gamma$ are pairwise co-prime.

Hence $x^3 + y^3 = \lambda^3\alpha\beta\gamma = z^3$, so

$\alpha\beta\gamma = \left(\dfrac{z}{\lambda}\right)^3$.

It follows from unique factorization and the fact that $\alpha,\beta,\gamma$ are pairwise co-prime that each of $\alpha,\beta,\gamma$ must be a perfect cube in $\Bbb Z[\omega]$.

Lemma 4: In the above, $\alpha + \beta + \gamma = 0$.

Proof: $\alpha + \beta + \gamma = $

$\dfrac{x+y}{\lambda} + \dfrac{x\omega + y\omega^2}{\lambda} + \dfrac{x\omega^2 + y\omega}{\lambda}$

$= \dfrac{x+y}{\lambda}(1 + \omega + \omega^2) = 0$.[/sp]

Another pause to catch our breath. We're getting pretty close, now, to deriving a contradiction.

 

[Deveno]

Re: Fermats last theorem (Not to freak, just a special one)

One last push, and our baby is born:

[sp]Suppose we have a minimal counter-example in $\Bbb Z[\omega]$ to Fermat's Last Theorem for $n$ = 3, that is a minimal non-trivial pairwise co-prime solution to:

$x^3 + y^3 = z^3$.

We have shown that there exist pairwise co-prime $\alpha,\beta,\gamma \in \Bbb Z[\omega]$ such that:

1.$\alpha\beta\gamma = \left(\dfrac{z}{\lambda}\right)^3$

2. $\alpha + \beta +\gamma = 0$.

3. Each of $\alpha,\beta,\gamma$ is a perfect cube in $\Bbb Z[\omega]$.

Let $k$ be the highest power of $\lambda$ that divides $z$. In the equation in (1), we therefore have $3k - 3 = 3(k-1)$ powers of $\lambda$ on the right-hand side.

However, from (2) and (3):

$\alpha + \beta +\gamma = x_0^3 + y_0^3 + z_0^3 = 0$

with $x_0,y_0,z_0$ pairwise co-prime, and this solution is also non-trivial (since $\Bbb Z[\omega]$ is a domain).

From an earlier corollary, we have $\lambda$ divides one of $x_0,y_0,z_0$.

Since our original solution is minimal, and $x_0,y_0,z_0$ are pairwise co-prime, $\lambda$ divides exactly ONE of these, and the LOWEST power of $\lambda$ that divides it is $k$ (or else our original solution is not minimal).

Thus in $\alpha\beta\gamma = (x_0^3)(y_0^3)(z_0^3)$, we have at least $3k$ powers of $\lambda$ and since $3k > 3k - 3$, we have our contradiction.

Therefore, no non-trivial pairwise co-prime solution exists, which is what we desired to actually prove.[/sp]

The astute reader will note this is actually an "infinite descent" proof in disguise.

********

Historical footnote: it is believed by some that an adaptation of this proof for higher-power roots of unity was the "marvelous proof" Fermat actually had in mind (that wouldn't fit in the margin). It is a reasonable conjecture, and the failure of cyclotomic integers to have unique factorization was not discovered until much later, perhaps forming the impetus for much of what we now call ring theory.

 

[mathbalarka]

Mess is in the eyes of the beholder. Meh.

 

[MarkFL]

Mess is in the eyes of the beholder. Meh.

 

[mathbalarka]

:D

Deveno, I am thinking of a approach through elliptic curves. Want more mess? :D

 

[Deveno]

Go for it! You might want to define some of your terms along the way :P

 

[mathbalarka]

Sure. I will use Mordells theorem. (cohomology theory! Fun, fun!)(Devil)