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wu :: forums    riddles    putnam exam (pure math) (Moderators: Eigenray, Grimbal, Icarus, william wu, towr, SMQ)    x^3 - 6x and x^4 - 8x^2 « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: x^3 - 6x and x^4 - 8x^2  (Read 2744 times) ThudnBlunder Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 x^3 - 6x and x^4 - 8x^2   « on: Jul 18th, 2007, 6:38pm » Quote Modify Find all irrational numbers x such that both x3-6x and x4-8x2 are rational. IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: x^3 - 6x and x^4 - 8x^2   « Reply #1 on: Jul 18th, 2007, 9:00pm » Quote Modify 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)] = [423] = 1 3, which gives another 4 solutions (with r=4). . IP Logged fengman Newbie     Posts: 1 Re: x^3 - 6x and x^4 - 8x^2   « Reply #2 on: Jul 20th, 2007, 4:38pm » Quote Modify How  do you apply euclidean algorithm? IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: x^3 - 6x and x^4 - 8x^2   « Reply #3 on: Jul 20th, 2007, 10:33pm » Quote Modify 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. IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: x^3 - 6x and x^4 - 8x^2   « Reply #4 on: Jul 20th, 2007, 11:13pm » Quote Modify 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. IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs => putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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