for simplicity I will use the following notation
xn should be taken to mean x^n
(1) x3+y=12x+20
(2) 2y3+z=24y+36
(3) 3z3+x=36z+52
solve (1) for y we get
y=12x+20-x3
putting this into (2) and solving for z we get
z=2x9-72x7-120x6+864x5+2880x4-1080x3-17280x2-28512x-15484
putting this into (3) moving everything to one side and then factoring we get (x-4)(a really big 26 degree polynomial I really don't feel like writing out :))=0
so we get x=4 for one solution which then gives us
z=4 and y=4
so (4,4,4) is a solution now all that is left is finding if there is any real zeros to that really big 26 degree polynomial.
To do this we shall analize the behavior of this polynomial. Let us call this polynomial F(x). If anyone really wishes to see this polynomial please post a request and when I get some more time later I will type it in.
First I note that F(x) is of even degree and since complex roots come in conjugate pairs it is possible for it to have no real roots (which is my conjecture). Next I will look at its end behavior. The leading term is 24x^6 and thus F(x) goes to +∞ at both ends. So once again this supports the possibility of no real roots. Now since both extremes are positive then for there to be no real zeros of F(x) then F(x)>0 for all x. So now we need to find all turning points F(x). Now if F(x) is positive at all of these turning points then F(x) must be positive for all x and thus no real zeros. Define the first dirivative of F(x) as F1(x). Now the turning points are found by solving F1(x)=0. Now F1(x) is a 25 degree polynomial and has the following factors
45(x+2)
(x^3-12x-22)
(x9-36x7-60x6+432x5-1440x4-540x3-8640x2-14256x-7741)
(13x12+24x11-528x10-1780x9+6300x8+37440x7+10500x6-265392x5-551520x4+115628x3+1730370x2+2257656x+983350)
(told you it was big :-) )
now I will consider each factor in turn. First finding any real zeros and if there are any finding if F(x) is positive at those zeros.
First factor gives us x=-2 and F(-2)=1 and thus is positive.
next factor has only 1 real zero and F(x) is positive at this zero as well
next factor has 3 real zeros and F(x) is positive for all of them as well
final factor has 4 real roots and F(x) is positive for all of them as well
Thus F(x) is positive at all of its turning points and thus F(x) is positive for all x.
So now after all this, we can finally conclude that (4,4,4) is the only solution :-)
Edited on July 30, 2006, 11:40 am
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Posted by Daniel
on 2006-07-30 11:36:17 |
solution with proof :-)
