Computer Science Canada Quick math question...for 500 bits

 Author: Martin [ Thu Jun 26, 2003 7:56 pm ] Post subject: Quick math question...for 500 bits X^n + Y^n = Z^n Find one case, with X,Y,Z and n integers, and n greater than 2 for which this is true.

 Author: Homer_simpson [ Thu Jun 26, 2003 8:01 pm ] Post subject: i guess if x,y and z were 0 and n was bigger than 2 it would be true...

 Author: Catalyst [ Thu Jun 26, 2003 8:23 pm ] Post subject: first off this is Fermat's last theorem and it has been proven to that is is not true second of all, homer since it is X,Y, and Z you should assume they are different numbers

 Author: AsianSensation [ Thu Jun 26, 2003 8:49 pm ] Post subject: aww....come on now martin, Fermat's Last Theorem? I would've expected something less conspicuous.....(doesn't really matter though, Catalyst or Bugz would probably get it anyways if it were a different question)

 Author: Homer_simpson [ Thu Jun 26, 2003 9:27 pm ] Post subject: oh ic... well in that case it is impossible i think

 Author: Andy [ Fri Jun 27, 2003 9:34 am ] Post subject: martin, quick my ass...

 Author: AsianSensation [ Fri Jun 27, 2003 9:49 am ] Post subject: yeah, didn't it take that guy(was it Andrew Wales?) like 8 years to come up with the 130 something pages proof for Fermat's Last Theorem?

 Author: Andy [ Fri Jun 27, 2003 12:18 pm ] Post subject: is it was actually proved?

 Author: Martin [ Fri Jun 27, 2003 1:18 pm ] Post subject: heh, some people have too much time

 Author: SilverSprite [ Fri Jun 27, 2003 2:35 pm ] Post subject: Andrew Wiles asian.... didnt mr white talk to you about him? he gave us a story last year about one of his students going crazy cuz he thought he had the proof..

 Author: AsianSensation [ Fri Jun 27, 2003 2:40 pm ] Post subject: i never heard Mr.White talk about him, cause im not in his class. lol, and to go crazy over some math proof.....i can only imagine two people right now, that would do that(SW and TTL), but then again, proving fermat's last theoreom guarentees \$50,000(or was it \$500,000?)

 Author: Andy [ Fri Jun 27, 2003 5:22 pm ] Post subject: i think TTL would juss jump of a bridge if she proved it

 Author: Crono [ Sat Jun 28, 2003 7:37 pm ] Post subject: would u give me 5000 bits if i proved fermat's last theorem rite here?

 Author: Andy [ Sun Jun 29, 2003 1:37 pm ] Post subject: lol ya he can prolly prove it too

 Author: SilverSprite [ Sun Jun 29, 2003 2:10 pm ] Post subject: yeah.. pshaaw what? like its hard??

 Author: bugzpodder [ Thu Jul 03, 2003 11:19 am ] Post subject: (-1)^3+(1)^3=0^3 so (-1,1,0) is a solution to x^3+y^3=z^3 did somebody say 500 BITS?!?!

 Author: bugzpodder [ Thu Jul 03, 2003 11:22 am ] Post subject: btw, no sneaking and changing the question!!

 Author: SilverSprite [ Thu Jul 03, 2003 1:25 pm ] Post subject: LMAO... ahahaha but i havent seen darkness around lately..

 Author: AsianSensation [ Thu Jul 03, 2003 1:26 pm ] Post subject: lol, good one Bugz, that's funny

 Author: bugzpodder [ Thu Jul 03, 2003 1:34 pm ] Post subject: yeah if anyone one is interested, pick any integer k and any positive p, (-k,k,0,2p+1) or (k,-k,0,2p+1) would work that is (a,b,c,n)

 Author: bugzpodder [ Thu Jul 03, 2003 5:58 pm ] Post subject: Its about time i get some bits! 500 sounds nice

 Author: SilverSprite [ Thu Jul 03, 2003 6:19 pm ] Post subject: bugzpodder wrote:yeah if anyone one is interested, pick any integer k and any positive p, (-k,k,0,2p+1) or (k,-k,0,2p+1) would work that is (a,b,c,n) well yeah thast sortof obvious.. the exponent is odd and so the two terms cancel each other out and give you 0.

 Author: bugzpodder [ Thu Jul 03, 2003 7:42 pm ] Post subject: so? your point being? i am simply stating the obvious! -|:o)

 Author: SilverSprite [ Thu Jul 03, 2003 9:54 pm ] Post subject: exactly you made it seem so complicated..

 Author: bugzpodder [ Fri Jul 04, 2003 11:22 am ] Post subject: we all know you are very bright silversprite.

 Author: SilverSprite [ Fri Jul 04, 2003 1:58 pm ] Post subject: akh.. meanie

 Author: Archi [ Mon Jul 28, 2003 11:32 am ] Post subject: 3^2+4^2=5^2 9+16=25 There we go...

 Author: bugzpodder [ Mon Jul 28, 2003 1:12 pm ] Post subject: n is greater than 2

 Author: SilverSprite [ Mon Jul 28, 2003 2:21 pm ] Post subject: lol

 Author: Archi [ Mon Jul 28, 2003 4:54 pm ] Post subject: He didn't say that it couldn't be 2.

 Author: SilverSprite [ Mon Jul 28, 2003 4:56 pm ] Post subject: Re: Quick math question...for 500 bits Darkness wrote:X^n + Y^n = Z^n Find one case, with X,Y,Z and n integers, and n greater than 2 for which this is true.

 Author: thegoose [ Fri Jun 18, 2004 5:29 am ] Post subject: It's funny that if you make X,Y,Z polynominals of degree greater than 1, the proof becomes about a page in length. I'll post it for 200 bits.

 Author: bugzpodder [ Fri Jun 18, 2004 8:35 pm ] Post subject: i dont understand what you mean... Quote: X,Y,Z polynominals of degree greater than 1 are you saying that there exists no polynomials f,g,h of degree greater than 1 such that (f(x))^n + (g(x))^n = (h(x))^n where n is an integer greater than 2? and let me ask if it is your own proof

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