### Monday, August 30, 1999

## JSH: FLT update, and an apology

Along with the discussions on how I use the term "factor" the weekend saw a post from Keith Ramsay that showed I hadn't covered every odd p>=5 for my Case 1 proof. Ok, I went ahead and fixed that today and the website is updated.

I apologize for getting a bit testy about the "factor" issue and jumping the gun as to how resolved everything is with my proof. I tend to jump to conclusions quickly and then I'm often forced to backtrack.

There is no doubt that everything has to be completely clear before anyone else is going to accept that I have a proof of FLT.

Now, back to the "factor" issue. My favorite explanation is this:

If you have f_1 + f_2 + ... + f_n = a, where 'a' is an integer and all the f numbers are nonreal and you then have

mf_1 + mf_2 +...+ mf_n = ma, and 'm' is an integer,

then of course, 'm' is a factor of ma.

All I've done is have something like

k_1 + k_2 +...+ k_n, where k = fa,

instead of writing out mf_1 + mf_2 +...+ mf_n

and said that each k has a "factor" of 'a'.

So, to take this out of the argument, all I'd have to do is write out, say,

mf_1 instead of k_1.

BUT, in my proof the expressions are a bit more complicated, so I'd actually have to write something like

(fgh)^p (f_3...f_(p-1)), and (fgh)^p(f-4...f_(p-1)) and somehow indicate that

the first doesn't have f_4 in it, while the second doesn't have f_3 in it.

Now, anyone can see that each of them *do* have (fgh)^p in them.

Of course, with complex numbers there's always the possibility when I multiply all these nonreals together and add them up they could do weird things like divide off some portion of the prime factors of (fgh) or multiply in more.

BUT, I cover this by showing that all the nonreals multiply to give a number q^p which has

So, that would still bring up the possibility that I'd screwed up the modular arithmetic somewhere in the actual proof page on my website. BUT, the way the proof works even if I did have some minor screw up, it still wouldn't change the fact that this method leads to a contradiction.

That's what's so incredibly powerful about it.

So, now I'm going to give my website for what I hope is the last time this will be necessary. If it does become known that I have a proof, I suspect it will become unavailable almost immediately. But at this rate, it could still be up for months.

http://www.mindspring.com/~jstev/FLTb.htm

I apologize for getting a bit testy about the "factor" issue and jumping the gun as to how resolved everything is with my proof. I tend to jump to conclusions quickly and then I'm often forced to backtrack.

There is no doubt that everything has to be completely clear before anyone else is going to accept that I have a proof of FLT.

Now, back to the "factor" issue. My favorite explanation is this:

If you have f_1 + f_2 + ... + f_n = a, where 'a' is an integer and all the f numbers are nonreal and you then have

mf_1 + mf_2 +...+ mf_n = ma, and 'm' is an integer,

then of course, 'm' is a factor of ma.

All I've done is have something like

k_1 + k_2 +...+ k_n, where k = fa,

instead of writing out mf_1 + mf_2 +...+ mf_n

and said that each k has a "factor" of 'a'.

So, to take this out of the argument, all I'd have to do is write out, say,

mf_1 instead of k_1.

BUT, in my proof the expressions are a bit more complicated, so I'd actually have to write something like

(fgh)^p (f_3...f_(p-1)), and (fgh)^p(f-4...f_(p-1)) and somehow indicate that

the first doesn't have f_4 in it, while the second doesn't have f_3 in it.

Now, anyone can see that each of them *do* have (fgh)^p in them.

Of course, with complex numbers there's always the possibility when I multiply all these nonreals together and add them up they could do weird things like divide off some portion of the prime factors of (fgh) or multiply in more.

BUT, I cover this by showing that all the nonreals multiply to give a number q^p which has

**no**prime factors in common with f, g or h.So, that would still bring up the possibility that I'd screwed up the modular arithmetic somewhere in the actual proof page on my website. BUT, the way the proof works even if I did have some minor screw up, it still wouldn't change the fact that this method leads to a contradiction.

That's what's so incredibly powerful about it.

So, now I'm going to give my website for what I hope is the last time this will be necessary. If it does become known that I have a proof, I suspect it will become unavailable almost immediately. But at this rate, it could still be up for months.

http://www.mindspring.com/~jstev/FLTb.htm