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        riddles >> putnam exam (pure math) >> x^3 - 6x and x^4 - 8x^2
        
(Message started by: ThudanBlunder on Jul 18th, 2007, 6:38pm)
Title: x^3 - 6x and x^4 - 8x^2
Post by ThudanBlunder on Jul 18th, 2007, 6:38pm
Find all irrational numbers x such that both x3-6x and x4-8x2 are rational.
Title: Re: x^3 - 6x and x^4 - 8x^2
Post by Eigenray on Jul 18th, 2007, 9:00pm
Suppose x3 - 6x - r = 0, and x4 - 8x2 - s = 0.

[hideb]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=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif6, which gives two solutions (with s=-12).  If s=-4, then we can solve for x = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(16+s)] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3, which gives another 4 solutions (with r=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif4).[/hideb].
Title: Re: x^3 - 6x and x^4 - 8x^2
Post by fengman on Jul 20th, 2007, 4:38pm
How  [hide]do you apply euclidean algorithm?[/hide]
Title: Re: x^3 - 6x and x^4 - 8x^2
Post by Eigenray on Jul 20th, 2007, 10:33pm
[hideb]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.)
[/hideb]

Of course, the result [hide](r2-24-2s)x - r(s+4)=0[/hide] may be easily verified just by expanding it, but the above explains how to derive it.
Title: Re: x^3 - 6x and x^4 - 8x^2
Post by Eigenray on Jul 20th, 2007, 11:13pm
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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