Friday, January 29, 1999


JSH: FLT conquered.

There's nothing like a little competition to stir the blood and get one excited about starting each day. I've been priveliged to be able to compete against the mathematicians of the world in the search for something most of them say doesn't exist.

Now that's the kind of competition I like, when your competitors deny there's actually a game to be played.

Whatever they may have believed....Time has brought the need for a change.

And I am the avatar of Change.


The proof is that for x^3 + y^3 = z^3, x,y,z can't all be nonzero integers.

Specifically, I consider x^3 + y^3 = z^3 with x,y,z integers as a special case of

X= F^3 + Fk; Z = F^3 + y; X^3 + y^3 = Z^3; k,y nonzero Integers;

X, F and Z are determined by k,y and the equations given.

Then an integer root f of F, requires than an x^3 + y^3 = z^3; x,y,z relatively coprime and nonzero exists.

Doing the substitutions for X and Z gives

(F^3 + Fk)^3 + y^3 = (F^3 + y)^3 expanding

F^9 +3F^6(Fk)+3F^3(Fk)^2 + (Fk)^3 + y^3 = F^9 + 3F^6 y + 3F^3 y^2 + y^3

subtracting F^9 + y^3 from both sides and collecting to the left gives

3F^7 k - 3F^6 y + 3F^5 k^2 + F^3 k^3 - 3F^3 y^2 = 0

dividing off 3F^3 k (since F = 0 is a trivial solution) gives

F^4 - F^3 (y/k) + F^2 k + (k^3 - 3y^2)/3k = 0

which has four roots for F. If FLT is false for p=3 then one of these roots is rational from the following:

with x^3 + y^3 = z^3; x,y,z relatively coprime and nonzero,

Now x^3 = (z-y)(z^2 + zy + y^2), which is an equation easily verified.

If z is divisible by 3 (one of them has to be,

Proof: Consider that x^3 + y^3 = z^3 means that (x+y-z)^3 = 3(z-x)(z-y)(x+y) which is easily verified by multiplying out the terms.

If neither x,y nor z are divisible by 3, then x^3 + y^3 = z^3 requires that x=y(mod 3) and x= - z (mod 3), but this means that none of (z-x)(z-y)(x+y) are divisible by 3, but the left side is a cube.)

then z = y + f^3;


Using the substitution z = y+a, with x^3 + y^3 = z^3,

x^3 + y^3 = (y+a)^3 so

x^3 = 3ay^2 + 3a^2 y + a^3

using a = z-y, I have

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

so z-y shares all of its prime factors with x and x^3/(z-y) has no prime factors in common with (z-y) or it would share them with y (remember x is not divisible by 3).

So, z-y = f^3 where f has to be an integer, since y and z are.

Then, since x^3 = (z-y)(z^2 + zy + y^2)

x^3 = f^3(z^2 + zy + y^2)

Using the result from before that division of x^3 by f^3 leaves no prime factors of f,

z^2 + zy + y^2 = m^3 where m must be an integer.

Now using z = y + f^3 to eliminate z.

z^2 + zy + y^2 = f^6 + 3f^3 y + 3y^2 = m^3

Since m is an integer there exists some integer k, such that

m = f^2 + k,

Substituting to eliminate m gives

f^6 + 3f^3 y + 3y^2 = (f^2 + k)^3,

expanding, the right and subtracting f^6 from both sides and bringing the other terms over to the right gives

3f^4 k - 3f^3 y + 3f^2 k^2 + k^3 - 3y^2 = 0;

Notice this requires that k be divisible by 3. Now dividing by 3k gives

f^4 - f^3 (y/k) + f^2 k + (k^3 - 3y^2)/3k = 0

Proving that this f is a root of

F^4 - F^3 (y/k) + F^2 k + (k^3 - 3y^2)/3k = 0.

Of course, it's also possible to consider this root 'f' with an integer 'y' and see what values of 'k' this allows.

Rewriting the equation gives

k^3 + 3f^2 k^2 + 3f^4 k - 3f^3 y - 3y^2 = 0;

k^3 + 3f^2 k^2 + 3f^4 k - 3y(f^3 + y) = 0;

using the same argument that I gave to show that z-y = f^3; z-x = g^3;

Because z must then be divisible by 3, so x+y = h^3/3;

Then for its integer value, k = gh; from k^3 = 3(Z-X)(X+y);

The other roots for k in the polynomial given must imaginary. This is obvious because with changing k, y and z remain the same and the equation is

x^3 + y^3 = z^3

Then the sum of all the roots for k, is 3f^2. The imaginary portion of the other roots cancel each other out in the sum, so the sum of the other two roots is

3f^2 + gh;

Therefore, the other two roots come from 3f^2/2 + gh/2 +/- imaginary stuff.

BUT, (f^2 + k)^3 = f^6 + 3f^3 y + 3y^2 = (f^2 + k)^3

from before and on the right there is no factor of 2, only integers.

if f,g and h are integers, one of them must be divisible by 2 and this would introduce a divisor of 2 on the left that isn't on the right, and the contradiction is reached.

The general proof works exactly the same way.


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