Monday, March 16, 1998
Proof of FLT? You be the judge.
Why try? Why not. I've been looking over what should have been old ground for a simple proof because Fermat said it was there; my intuition says it's there; and because I've felt like it (it has been fun most of the time).
Problem:
Given x^p + y^p = z^p; x,y,z relatively prime; p odd prime; show no solution exists.
My current method (which hasn't been disproven as of yet) is to take
(s+a)^p + (s+b)^p = (s+c)^p
as a subs. for x^p + y^p = z^p.
Then I choose a,b,c such that a+b=c which interestingly means that
s = x+y-z. [(a=z-y); (b=z-x); (c=2z-(x+y))]
I then expand and group and also notice that a^p + b^p - c^p = -pabcQ (Q is just an integer to handle the general expansion, for p=3, Q=1).
I then note that the eqn thus found must be of the form (s+A)(s+B)(s+C)...=0
Finally, I notice that c=2z-(x+y)=z + z-(x+y) = z - s, and that this reduces the eqn thus formed from a highest s of s^p to s^{p-1} (you have to be working with the expansion to see this)
But notice that the one of the roots must be divisible by z and I have a contradiction because the roots of the original aren't.
The above shouldn't make a lot of sense because I'm in a bit of a hurry and am pushing out the conclusion. I would recommend that you do two things to get an idea of what I'm doing.
1. Try it with x^2 + y^2 = z^2 (which gives s^2 + a^2 + b^2 - c^2 = 0, note that
(a+b-c)^2 = a^2+b^2-c^2 + 2ab)
2. Then try it with p=3 because the expansions are simpler like
(a+b-c)^3 = a^3+b^3 - c^3 + 3abc
The method is a bit strange even to me but it seems to work. It leaves the question to later (once it has been verified to work by others) as to why it wasn't found earlier.
Get through the above and you will be one of the few people who currently know the simple explanation of why the theorem is true. After all, Wile's proof is a bit complex.
Problem:
Given x^p + y^p = z^p; x,y,z relatively prime; p odd prime; show no solution exists.
My current method (which hasn't been disproven as of yet) is to take
(s+a)^p + (s+b)^p = (s+c)^p
as a subs. for x^p + y^p = z^p.
Then I choose a,b,c such that a+b=c which interestingly means that
s = x+y-z. [(a=z-y); (b=z-x); (c=2z-(x+y))]
I then expand and group and also notice that a^p + b^p - c^p = -pabcQ (Q is just an integer to handle the general expansion, for p=3, Q=1).
I then note that the eqn thus found must be of the form (s+A)(s+B)(s+C)...=0
Finally, I notice that c=2z-(x+y)=z + z-(x+y) = z - s, and that this reduces the eqn thus formed from a highest s of s^p to s^{p-1} (you have to be working with the expansion to see this)
But notice that the one of the roots must be divisible by z and I have a contradiction because the roots of the original aren't.
The above shouldn't make a lot of sense because I'm in a bit of a hurry and am pushing out the conclusion. I would recommend that you do two things to get an idea of what I'm doing.
1. Try it with x^2 + y^2 = z^2 (which gives s^2 + a^2 + b^2 - c^2 = 0, note that
(a+b-c)^2 = a^2+b^2-c^2 + 2ab)
2. Then try it with p=3 because the expansions are simpler like
(a+b-c)^3 = a^3+b^3 - c^3 + 3abc
The method is a bit strange even to me but it seems to work. It leaves the question to later (once it has been verified to work by others) as to why it wasn't found earlier.
Get through the above and you will be one of the few people who currently know the simple explanation of why the theorem is true. After all, Wile's proof is a bit complex.
# posted by James Harris @ 18:58 
