Comments on: Metamath

By: knuddavid

knuddavid — Sun, 16 Jul 2006 18:29:14 +0000

Bear with me, I am thoroughly embarrassed. I believe my problems stem from the fact that I tried to copy a MSWord document, and the control characters are getting in the way.
I have now copied it in HTML format and done my best to edit the result:

Hesselink asks:
Would it not be easier to find a simpler proof of Fermatâ€™s Last Theorem, formalize that, and verify that by computer?

The problem:
(1) a^n + b^n = c^n
This equation has integer solutions if â€˜n = 1â€™, or â€˜n = 2â€™; but not for any other integer value of â€˜nâ€™. It is evidently false for â€˜n = 0â€™, trivially true for â€˜n = 1â€™, and known to have solutions for â€˜n = 2â€™. How would he go about proving it to be false for any higher number of â€˜nâ€™, given the constraint that only integer arithmetic was allowed?
The expression arose from a problem in geometry, and it is my guess that Fermat turned to the study of triangles for proof.
Stipulation:

(2) 0 2, for a solution to satisfy Eq.(1):

(5) c^n = (a^2 +b^2 – 2abcosC )n/2 = a^n + b^n + Rn

Excluding the special case: cosC = 0; with a,b,c being Pythagorean triples, if

cosC (a^n +b^n ) for all values n 0).

When cosC moves into positive territory, the landscape changes. We have to rephrase Stipulation(2) to ensure that c remains the largest side of the triangle. Since

(6) a = bcosC + ccosB

a necessary condition for the current discussion is

(7) a > (b + c) cosC.

As cosC increases in value, the process ends in the frozen stalemate of the isosceles triangle, when cosC = 0.5, and a = b = c. We then have c^n

By: knuddavid

knuddavid — Sun, 16 Jul 2006 17:18:20 +0000

Sorry, this is not my day. In plain text:

Rn is zero only in the interval of cosC between 0 and 0.5.

In Eq.(3), if c is integer, its square must be integer, and the polynomial must be in quadratic form. This requires cosC = +/-1 in violation of the conditions for a solution of Eq.(1).

(I hope it works this time)

By: knuddavid

knuddavid — Sun, 16 Jul 2006 17:06:36 +0000

Once more, before I give up:

We then have c^n

By: knuddavid

knuddavid — Sun, 16 Jul 2006 16:59:49 +0000

Well, I try again. I hope the editor can splice it together:

As cosC increases in value, the process ends in the frozen stalemate of the isosceles triangle,
when cosC = 0.5, and a = b = c.

We then have c^n

By: knuddavid

knuddavid — Sun, 16 Jul 2006 16:56:31 +0000

I am new to this procedure, and shall continue where I was interrupted:

As cosC increases in value, the process ends in the frozen stalemate of the isosceles triangle,
when cosC = 0.5, and a = b = c.

We then have c^n

By: knuddavid

knuddavid — Sun, 16 Jul 2006 16:50:55 +0000

Hesselinks asks:
Would it not be easier to find a simpler proof of Fermat’s Last Theorem, formalize that, and verify that by computer?

I don’t pretend to have a proof but I suggest a different approach that Fermat himself (not having an electronic computer) might have tried.

The problem:

(1) a^n + b^n = c^n

This equation has integer solutions if â€˜n = 1â€™, or â€˜n = 2â€™; but not for any other integer value of â€˜nâ€™. It is evidently false for â€˜n = 0â€™, trivially true for â€˜n = 1â€™, and known to have solutions for â€˜n = 2â€™.
How would he go about proving it to be false for any higher number of â€˜nâ€™, given the constraint that only integer arithmetic was allowed?

The expression arose from a problem in geometry, and it is my guess that Fermat turned to the study of triangles for proof.

Stipulation:

(2) 0 2, for a solution to satisfy Eq.(1):

(5) c^n = (a^2 +b^2 – 2abcosC)^(n/2) = a^n + b^n + Rn

Excluding the special case: cosC = 0; with a,b,c being Pythagorean triples, if cosC (a^n +b^n ) for all values n > 2, (i.e. Rn > 0).

When cosC moves into positive territory, the andscape changes.
We have to rephrase Stipulation(2) to ensure that c remains the largest side of the triangle.

Since

(6) a = bcosC + ccosB

a sufficient condition for the current discussion is

(7) a > (b + c) cosC.

As cosC increases in value, the process ends in the frozen stalemate of the isosceles triangle, when

cosC = 0.5, and a = b = c.

We then have c^n

By: PeterMcB

PeterMcB — Wed, 17 Aug 2005 08:42:09 +0000

Thanks. Problem 2 (a computer to play GO) is definitely a major problem, but the others are not IMHO. Perhaps I should have expected a *theoretical* computer scientist to ignore what I think are the major challenges in the discipline (modulo my Dutch translation).

By: sigfpe

sigfpe — Tue, 16 Aug 2005 18:03:40 +0000

Here you go. Page 53. But there's a slight catch...

By: PeterMcB

PeterMcB — Tue, 16 Aug 2005 09:58:51 +0000

Hesselink’s Fermat page mentions this prooblem as a challenge in theoretical computer science, part of a set recently articulated by Prof. dr. Jan Bergstra. As a computer scientist, I was interested to see this list of challenges, but had no success finding it. Do you happen to know if Bergstra’s Problems exist anywhere on the wetb?