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 ]
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i guess if x,y and z were 0 and n was bigger than 2 it would be true... Rolling Eyes

Author:  Catalyst [ Thu Jun 26, 2003 8:23 pm ]
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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 ]
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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 ]
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oh ic... well in that case it is impossible i think Confused

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

Author:  AsianSensation [ Fri Jun 27, 2003 9:49 am ]
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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 ]
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is it was actually proved?

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

Author:  SilverSprite [ Fri Jun 27, 2003 2:35 pm ]
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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 ]
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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 ]
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i think TTL would juss jump of a bridge if she proved it

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

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

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

Author:  bugzpodder [ Thu Jul 03, 2003 11:19 am ]
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(-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 ]
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btw, no sneaking and changing the question!!

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

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

Author:  bugzpodder [ Thu Jul 03, 2003 1:34 pm ]
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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 ]
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Its about time i get some bits! 500 sounds nice Very Happy

Author:  SilverSprite [ Thu Jul 03, 2003 6:19 pm ]
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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 ]
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so? your point being? i am simply stating the obvious! -|:o)

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

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

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

Author:  Archi [ Mon Jul 28, 2003 11:32 am ]
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3^2+4^2=5^2

9+16=25

There we go...

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

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

Author:  Archi [ Mon Jul 28, 2003 4:54 pm ]
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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 ]
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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 ]
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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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