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...
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.
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 ?? :crazy:
But 3^3 + (-2)^3 + (-1) ^3 = 27-8-1 = 18 ?? :crazy: Further, 3^2 + (-2)^2 + (-1)^2 = 9 + 4 + 1 = 14 ?! :crazy:
Oh, I see .. looks like shabbir referred to this thread (from Physics Forum) for the answers : http://www.physicsforums.com/showthread.php?t=315700. But, the correct answers are not 3, -2, -1. The answers are : x= 2 cos (20) y= 2 cos (140) z= 2 cos (260) Probably shabbir didn't read the last post carefully. It mentioned a nice short-cut method to solve, assuming X=3, Y=-2, Z=-1. But, they are not the real answers. Anyway, congrats xpi0t0s
This code found multiple results; it only searches for approximations (hence if test1>=-0.05 && test2<=0.05) but all the results revolve around 6 hence the guess that there was an integral solution that was precisely 6.0. But this would make x,y,z non-integral and therefore irrational. Code: void powers() { FILE *fp; fp=fopen("results.txt","a"); double x,y,z; for (x=1.5; x<=10.0; x+=0.00001) { for (y=-10.0; y<=10.0; y+=0.00001) { // don't need to loop over z; from x+y+z=0 we can calc z z=-(x+y); double test1=x*x*x+y*y*y+z*z*z-3.0; if (test1>-0.05 && test1<0.05) { double test2=x*x*x*x*x+y*y*y*y*y+z*z*z*z*z-15.0; if (test2>-0.05 && test2<0.05) { fprintf(fp,"Sum of squares of x,y,z=%f; x=%f y=%f z=%f\n",x*x+y*y+z*z,x,y,z); } } } } fclose(fp); } The closest results found were: Sum of squares of x,y,z=6.000000; x=1.879390 y=-1.532080 z=-0.347310 Sum of squares of x,y,z=6.000000; x=1.879390 y=-0.347310 z=-1.532080 and I'm still working on getting this in surd form which is proving difficult (for me), basically on paper what I did was to try to solve by simultaneous equations and use Gaussian elimination: [1] x+y+z=0 [2] x^3+y^3+z^3=3 [3] x^5+y^5+z^5=15 First step is to eliminate y from [2] and [3] [1]->y=-(x+z) So [2]->x^3-(x+z)^3+z^3=3 -> 3x^2+3zx^2=-3 [2a] and [3]-> x^5-(x+z)^5+z^5=15 [3a] So now we rearrange [2] as (3x)z^2+(3x^2)z+3=0 which gives us x in terms of x and we can use the standard quadratic solver (-b+/-sqrt(b^2-4ac))/2a whicg gives [4] z=-x/2 +/- sqrt((x^3-4)/4x) hence the inequality - for z to be real, x^3-4>=0 so x>cube root of 4 or x>=4^(1/3), which is why x is initialised in the program as x=1.5: 1.5<4^(1/3). So the next step in Gaussian elimination is to eliminate z from [3]. This equation becomes: [3a] 5x^4z + 10x^3z^2 + 10x^2z^4 + 5xz^4 = -15 and the next step (which I woke up with this morning) is to rearrange this as z=f(x) then to replace z with [4] and there's an equation purely in terms of x which can then be solved to give a range of values for x. I can't be arsed doing that on paper now so I've installed Maxima to see if I can do it with that.
Here we go, in installation time plus about 5 seconds: (%i1) solve([x+y+z=0,x^3+y^3+z^3=3,x^5+y^5+z^5=15],[x,y,z]); (%o1) [[x = - 0.34729635646322, y = 1.879385232208487, z = - 1.532088888888889], [x = - 1.532088888888889, y = 1.879385232208487, z = - 0.34729635646322], [x = - 0.34729635646322, y = - 1.532088888888889, z = 1.879385232208487], [x = 1.879385232208487, y = - 1.532088888888889, z = - 0.34729635646322], [x = - 1.532088888888889, y = - 0.34729635646322, z = 1.879385232208487], [x = 1.879385232208487, y = - 0.34729635646322, z = - 1.532088888888889]] and [x = 1.879385232208487, y = - 1.532088888888889, z = - 0.34729635646322] pretty much matches what I got.
