Harristotelian Logic

For a while now I've been talking about how you can actually test my proof of Fermat's Last Theorem.

Of course, the problem with that is that it can look a little hairy after you pass p=3.

So how do you test the proof with actual numbers?

Well, the proof requires that 2 have the factor of x^2 + y^2 - z^2 not shared with x and y as a factor, which keeps x, y and z from being integers (remember x^p + y^p = z^p).

Some of you may have noticed that for p=3, x=1, and y=1, you get

2^{1/3}/(2^{1/3} - 1), which is easily shown to not be a fraction.

Others may have looked at the gaussian integer solution that's been posted and noticed it worked as well, and it's a LOT easier to work with since x^2 + y^2 - z^2 gives an integer (though then x^2 + z^2 - y^2 is a bit of a pain).

But anyway, past p=3, the easiest example gets a good bit harder to evaluate.

After all, is 2^{(p-2)/p}/(2^{(p-2)/p} - 1) a fraction or not?

Strangely enough, or I'd think you'd know the proof is correct by now, your professors can't help you with that one, but the truth CAN be determined.

You can just use a quadratic ring bridge.

Using fx^2 + gx + h = (a1 x + b1)(a2 x + b2), and letting h = 2^{p-2}-1, b1 = 2^{(p-2)/p} - 1, and

a1 b2 = m, a2 b1 = 2, it's just a matter of finding m and f, such that

2 = (-(m+2) + sqrt((m+2)^2 - 4(2^{p-2}-1)f))/2

so

4 + (m+2) = sqrt((m+2)^2 - 4(2^{p-2}-1)f))

so

((m+2) + 4)^2 = (m+2)^2 - 4(2^{p-2}-1)f

so

f = -(((m+2) + 4)^2 - (m+2)^2)/(4(2^{p-2}-1))

so

f = -((m+2) + 4 - m - 2)((m+2) + 4 + m + 2)/(4(2^{p-2}-1))


so

f = 2(m + 4)/(2^{p-2}-1).

Of course, there's an integer m such that m+4 is divisible by 2^{p-2}-1.

And that's it.

I've proven that b2 IS indeed a factor of 2.

Note: I don't know what a1 is, and don't care.