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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Go4Expert Member
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| 18Jul2009,04:36 | #2 |
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Hey...
x^2+y^2+z^2 =1 Im i right??? |
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Mentor
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| 18Jul2009,05:42 | #3 |
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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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Mentor
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| 18Jul2009,06:15 | #4 |
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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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Go4Expert Founder
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| 18Jul2009,07:19 | #5 |
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Quote:
Originally Posted by xpi0t0s  I have the values of x,y,z as well if you want to. |
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Banned
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| 18Jul2009,07:48 | #6 |
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@ xpi0t0s, Congrs Buddy
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~ Б0ЯИ Τ0 С0δЭ ~
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| 18Jul2009,12:56 | #7 |
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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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~ Б0ЯИ Τ0 С0δЭ ~
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| 18Jul2009,12:59 | #8 |
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I wonder why I keep getting 60 as the answer.
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Go4Expert Founder
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| 18Jul2009,13:02 | #9 |
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The answer is 6 and X=3, y=-2 and Z=-1
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~ Б0ЯИ Τ0 С0δЭ ~
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| 18Jul2009,13:06 | #10 |
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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 ?!
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