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.
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.
# posted by James Harris @ 13:14 
