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Re: There are liars among you
Posted:
Aug 4, 2000 8:43 PM


On Thu, 03 Aug 2000 16:01:51 GMT, jastha@mydeja.com wrote, in part:
>When I post, I typically put in mathematical content.
>There are people who never do in reply.
Yes, I'm one of them. There are those, though, who have shown errors you have made in the past.
Your argument here is an invalid one, simply because there are many people who claim to have proved FLT or done something else amazing, and real mathematicians have better things to do than go through such claims and prove exactly where they're wrong. So the fact that they don't doesn't prove all these other people right.
>Answer this post with an actual math error from my post
>"For mathematicians who lack flexibility".
Well, I found a post titled "For mathematicians lacking flexibility"; will that do?
First you say:
>I take the expressions
> z^2 = kxy(mod (x+y+vz)/h)
> x+y+vz = 0(mod x+y+vz)
>and
> x^p + y^p = z^p.
Then:
>It's easy to handle h, since it's just
> (x+y)^{1/p} for Case 1 or (p(x+y))^{1/p} for Case 2.
And then:
>You see, x^p + y^p is divisible by x+y. So, since z^p = x^p + y^p, it >shares factors in common with x+y.
I do remember that x^2  y^2 equals x+y times xy. This is because the xy term cancels out.
Now, if x=2, y=1, and p=3, it's true that 8+1 is divisible by 3. If x=5 and y=3, with p still equal to 3, 125+27 is 152, and that is a multiple of 8. So without performing fancy polynomial divisions, at least I've checked and seen that this could be true.
>That leaves k and v.
>I start with v, and from v, I find out about everything else.
In other words, you are setting v= some expression "without loss of generality". Mathematicians do that all the time.
> x+y+vz = 0(mod x+y+vz), which as a tautology doesn't say anything at >this point, but it will be useful later,
> z^2 = kxy(mod (x+y+vz)/h), and finally
> x^p + y^p = z^p.
The rest is for p=3, and we do know that there _is_ a simple proof for the p=3 case of FLT. So there doesn't have to be an error in this post for your "proof of FLT" to be wrong! Not a single one.
> 3v = k(v^3 + 1)(mod (x+y+vz)/h).
You are then claiming that the obstacle to understanding your proof is:
>Of all these values can any of you accept that
> 3v = k(v^3+1)
>might just be included as one of them?
Well, if v and k are supposed to be integers, v^3 is usually bigger than 3v, unless v=1. And in that case, k is 3/2, which isn't an integer.
But maybe they don't have to be integers. There really isn't that much claimed in this post of yours to be able to find an "error" in it.
John Savard (teneerf <) http://home.ecn.ab.ca/~jsavard/crypto.htm



