Sunday, August 25, 1996


For the problem solvers, an approach to FLT

Since I'm not sure that last was brief enough I'm giving the necessary information to come to your own conclusion. There's the way I went and there's the direct way. I'll give the start to both.

Given x,y,z relatively prime naturals and n odd prime greater than 3 prove FLT through contradiction.

Note: (x+y), (z-x), (z-y) must have an nth root if not divisible by n and have an n^{n-1} factor with others with nth roots if divisible by n.

i.e. x+y=f^n or x+y=n^{n-1}f^n

Also, that x,y, or z must be divisible by 3 or there's a proof for the first case (I think it's called)

Here's what you do for the direct approach.

1> Use x + delta x = my, y + delta y = mx, z + delta y = mz

2> Solve the deltas in terms of x,y and z.

3> Take x^n + y^n = z^n and cube it.

4> Substitute the (x+delta x), (y+delta y), and (z+delta y) for x,y, z

5> Expand and subtract.

6> Now substitute in the values of the deltas in terms of x,y,z

7> Look at the denominator of the one term without a visible factor of 3 and prove that y can't be divisible by 3.

8> Consider that you can reverse your x and y.

9> Do the above up to 7> with the following

x + delta x = -mz, y + delta y = my, z + delta y = -mx

and show that z can't be divisible by 3.

10> Pat yourself on the back.

My way was to look at x^3 + y^3 = z^3. I then did the above but used the erroneous notion that (z-x) if divisible by 3 must have a factor of 3^3 (it actually must have a factor of 3^2)

With that erroneous assumption, prove FLT for n=3. Then look at the above again, and then pat yourself on the back. It's all in the 3's.


Minor correction and maybe why this is new, Re: FLT

As usual, almost immediately after that latest post I realized an error. I stated that (z-x) has an nth root. That's true unless it is divisible by n, then it has an (n-1)th root for n and an nth root for other factors, that is, (z-x) = n^n-1 f^n

That means that what I gave isn't a proof for n=3 but that doesn't bother me because it is correct for any othe odd prime. That's because in those cases it is an nth root.

That's probably why I didn't notice it before and why it wouldn't have been noticed until now. I had to make a mistake to get it because if I had remembered that, I would have given up again (and berated myself mercilessly).


Ok, I've lied but I need help again. Re:FLT

My last post was an apology for annoying you with my attempts at proving FLT with simple algebra which is obviously an impossible task, and such attempts have been a source of irritation for mathematicians for hundreds of years. I was forcing myself off because I was scared that I was insane and hopelessly obsessed with a Quixotic approach, and was just making a complete fool of myself.

However, I did have some reasons for hope. I had talked to several mathematicians previously and one recommended I contact the AMS. I did so and ended up (after some importuning behavior justified in the name of zealotry) corresponding with one of the top editors at the AMS. He trashed my conclusions but noted that my beginning and middle were correct. After bothering him enough, and not being satisfied with the information I obtained I finally started posting here. Eventually, I put out my method, found that no one else had ever seen any of it before, and got into another one on one with a math grad who helped trash my latest conclusions.

Not being able to let it go, I finally managed to stay underground long enough for my subconscious to have a go at it.

So, now I'm back with the hope that someone will be kind enough to critique it again. I promise to be much more polite this time :)

Rather than go over everything in detail again I hope someone will reply who saw the originals. For anyone else, I did post something on sci.physics as "A Physicist Tackles FLT" journeyman, with a subsequent post correcting a mistake in that original.

So, I'm going at it one more time because I admit I'm totally obsessed and I think I have something. Here's the ideas based on the assumption of an understanding of the method.

1> For a partial recap, x,y,z relatively prime, (z-x) has an nth root as does (z-y) and (x+y).

2> Either x,y or z has to be divisible by 3 or there's a proof of FLT in the first case (or the second I'm not sure).

3> I substitute in the original equation using

x + delta x = my, y + delta y = mx, z + delta y = mz

which determines all of my deltas and m in terms of x,y and z

4> Substituting (x + delta x) for x and the same for the others, expanding and subtracting off x^n + y^n = z^n, and inserting the explicit valules for the deltas, I get a strange looking equation with (z-x)'s in the denominator.

5> If I do this with n=3 I get the following

3x^2(x+y-z) + 3x(x+y-z)^2(x-y)/(z-x) + (x+y-z)^3(x-y)^2 = 3z(z^2-y^2)+ 3z^2(x-y)(z-y)/(z-x)

What I failed to notice previously is that this is the beginning of a simple proof for n=3. Note that if (z-x) is divisible by 3 it would have to have a factor of 3^3. But all terms except the last one on the left are multiplied by 3. The result is that that term has an extra factor of 3 in its denominator which makes it the only such term.

6> Since I can reverse x and y that proves that they can't be divisible by 3 which leaves z, but I can do another expansion using the following deltas

x + delta x = -mz, x + delta y = my, z + delta y = -mx

which forces the same conclusion for z.

7> So I had a proof for n=3 but so what. I kept wondering how to force 3's onto higher odd primes. And then it finally hit me, multiply.

(x^n + y^n)^3 = z^3n

x^3n + 3x^2n y^n + 3x^n y^2n + y^3n = z^3n

I leave it to anyone who cares to work out that the method works for the above. The only terms that need to be considered are those middle ones but notice— they all have 3's.

Such simplicity still astounds me although I expected it. What a fascinating subject.

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