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Topic: x^3 - 6x and x^4 - 8x^2 (Read 2775 times) |
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ThudnBlunder
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The dewdrop slides into the shining Sea
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x^3 - 6x and x^4 - 8x^2
« on: Jul 18th, 2007, 6:38pm » |
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Find all irrational numbers x such that both x3-6x and x4-8x2 are rational.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Eigenray
wu::riddles Moderator Uberpuzzler
    

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Re: x^3 - 6x and x^4 - 8x^2
« Reply #1 on: Jul 18th, 2007, 9:00pm » |
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Suppose x3 - 6x - r = 0, and x4 - 8x2 - s = 0. hidden: | Applying the Euclidean algorithm, we find that (r2-2s-24)x - r(s+4) = 0. So the only way x can be irrational, with r,s rational, is if we have r2-2s-24=0, and r(s+4)=0. Now either r=0 or s=-4. If r=0, then since x is irrational we must have x= 6, which gives two solutions (with s=-12). If s=-4, then we can solve for x =  [4 (16+s)] =  [4 2 3] = 1  3, which gives another 4 solutions (with r= 4). | .
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fengman
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Re: x^3 - 6x and x^4 - 8x^2
« Reply #2 on: Jul 20th, 2007, 4:38pm » |
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How do you apply euclidean algorithm?
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Eigenray
wu::riddles Moderator Uberpuzzler
    

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Re: x^3 - 6x and x^4 - 8x^2
« Reply #3 on: Jul 20th, 2007, 10:33pm » |
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hidden: | We want to find the gcd of the polynomials x4-8x2-s and x3-6x-r. So we perform division with remainder: x4-8x2-s - x*(x3-6x-r) = -2x2+rx-s x3-6x-r + x/2*(-2x2+rx-s) = r/2 x2 - (6+s/2)x - r -2x2+rx-s + 4/r*[r/2 x2-(6+s/2)x-r] = (r-24/r-2s/r)x - (s+4). Now, if x4-8x2-s and x3-6x-r are both 0, then everything above is 0, which gives the result. Another way of putting it is that the minimal polynomial of x divides both x4-8x2-s and x3-6x-r, and so divides their gcd, which must necessarily divide (r2-24-2s)x - r(s+4). (In the case of the solutions, this is actually the zero polynomial, so it is not actually the gcd.) | Of course, the result (r2-24-2s)x - r(s+4)=0 may be easily verified just by expanding it, but the above explains how to derive it.
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Eigenray
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Re: x^3 - 6x and x^4 - 8x^2
« Reply #4 on: Jul 20th, 2007, 11:13pm » |
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Carrying out the above a bit further shows that if r = x3-6x, s = x4-8x2, then s(12+s)2 = r4 - 32r2, independent of x. This leads to an interesting problem: Given polynomials f(x) and g(x), can we find polynomials F and G such that F(f(x)) = G(g(x))? Here, f(x) = x(x2-6), F(x) = y2(y2-32) g(x) = x2(x2-8), G(x) = x(x+12)2.
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