Math Riddle | 17 Jul 2009

shabbir's Avatar, Join Date: Jul 2004
Go4Expert Founder
A bit tricky and interesting as well

x+y+z=0
x^3+y^3+z^3=3
x^5+y^5+z^5=15

x^2+y^2+z^2=?

^ = Power of
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rik625's Avatar, Join Date: Jun 2009
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Hey...

x^2+y^2+z^2 =1

Im i right???
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xpi0t0s's Avatar, Join Date: Aug 2004
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It'd better not be as simple as that...after several hours and sheets of paper I'm just about to give up on solving it with simultaneous equations...
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xpi0t0s's Avatar, Join Date: Aug 2004
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However that did give me a useful inequality...so I'm going to register a guess, that x^2+y^2+z^2 is an integer and is precisely 6. Just running another, more accurate, test but it'll take a while to produce results.
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shabbir's Avatar, Join Date: Jul 2004
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Quote:
Originally Posted by xpi0t0s View Post
However that did give me a useful inequality...so I'm going to register a guess, that x^2+y^2+z^2 is an integer and is precisely 6. Just running another, more accurate, test but it'll take a while to produce results.
Yes that is the right Answer

I have the values of x,y,z as well if you want to.
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naimish's Avatar
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@ xpi0t0s, Congrs Buddy
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SaswatPadhi's Avatar, Join Date: May 2009
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Of course X^2+y^2+z^2 can be real even if x, y, z are all complex. But, I'm just curious .. Do these equations have any real solutions ??
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SaswatPadhi's Avatar, Join Date: May 2009
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I wonder why I keep getting 60 as the answer.
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shabbir's Avatar, Join Date: Jul 2004
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The answer is 6 and X=3, y=-2 and Z=-1
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SaswatPadhi's Avatar, Join Date: May 2009
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But 3^3 + (-2)^3 + (-1) ^3 = 27-8-1 = 18 ??

Further, 3^2 + (-2)^2 + (-1)^2 = 9 + 4 + 1 = 14 ?!