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May 16th, 2011, 01:14 AM  #1 
Newbie Joined: May 2011 Posts: 13 Thanks: 0  Solving equation in rational roots
Hi, I have an equation of the form ax1 + bx2 + cx3 = d, where: x1 = ((3sqrt(5))/2)^1/2 x2 = x1^2 x3 = x1^3 a,b,c, and d are rational numbers. I need to know whether it is possible to solve this equation in rational numbers. Thanks for your help! 
May 16th, 2011, 05:06 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Solving equlation in rational roots
Of course the origin is a trivial solution. As it happens, all of your x_i are of the form q_1 + q_2 * sqrt(5), where q_1 and q_2 are rational. So there should be infinitely many solutions in the rationals (choose anything for a and b, then choose the c that makes the sum of the coefficients of sqrt(5) equal to zero). 
May 16th, 2011, 12:32 PM  #3 
Newbie Joined: May 2011 Posts: 13 Thanks: 0  Re: Solving equation in rational roots
In fact, I was trying to the following issue 9found in my other post 'Number of closed paths formed by arcs of one fifth of a circle': Find the number of possible closed paths using one fifth of an arc (72 degrees), where at each time step we can move either clockwise or anticlockwise. in that particular problem, it was assumed that the number of arcs should be 70. In my solution, I assumed that all closed paths have no loops. I then proved that the number of arcs should necessarilly be even in order to get a closed path. Then, I simplified the problem to finding the number of possible EQUILATERAL POLYGONS, with interior angles: 72, 2*72, 3*72, and 4*72. To do this I needed to find the the combinations of interior angles satisfying the following conditions: 1 Sum of cosines of the interior angles = (n/2), where n is the number sides of the polygon, which in turn is fixed and directly related to the number of arcs. 2 Sum of the interior angles = (n2)*180 (equation 1) I then assumed that x,y,u, and v are the number of angles of size 72, 2872, 3*72, and 4*72 respectively. These satisfy the following equations: 1 x+y+u+v = n 2 x+2y+3u+4v = (5/2)*(n2) Thus, I was left only with 2 unknowns: x and y! After subsituting x and y in (equation 1) and doing sum math, and also noting that cos(72) = sqrt((3sqrt(5))/2), the problem was down to the following: Finding the solution of the equation: A(1,x,y,n)cos(72) + B(1,x,y,n)[cos(72)]^2 +C(1,x,y,n)[cos(72)]^3 = D(x,y,n) where A, B, and C are linear functions of 1,x,y, and n and D is a linear function of x,y, and n. But I was unable to continue. I wonder whether we can assume that the powers of cos(72)( are linearly independent in the ring of rational numbers, and use this fact to find the unknowns. Thanks! 
May 16th, 2011, 12:47 PM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Solving equation in rational roots Quote:
They are not, as suggested by my post above. They are all of the form a + b*sqrt(5), where a and b are rational.  
May 16th, 2011, 12:55 PM  #5 
Newbie Joined: May 2011 Posts: 13 Thanks: 0  Re: Solving equation in rational roots
Yes, 72 degrees. Sorry, but I did not understand how can cos(72) = sqrt((3sqrt(5))/2) be written in the form: a + b*sqrt(5), where a and b are rationals. Or cos(72)^3 = ((3sqrt(72))/2)*sqrt((3sqrt(5))/2). 
May 16th, 2011, 09:23 PM  #6  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Solving equation in rational roots Hello, rhseiki! Quote:
 
May 16th, 2011, 10:31 PM  #7 
Newbie Joined: May 2011 Posts: 13 Thanks: 0  Re: Solving equation in rational roots
Hi soroban, Exactly! We assume that x,y, and z are given, and we need to find the rational solutions of the set of equation that u mentioned. In fact, these set equations have nontrivial solutions (otherwise, the would not be able to have various equilateral polygons with angles multiples of 72 degrees), so I just need a method to explicitely find a,b,c, and d. 
May 17th, 2011, 05:04 AM  #8  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Solving equation in rational roots Quote:
x = ((3sqrt(5))/2)^1/2 1/x = (2/(3sqrt(5)))^1/2 1/x^2 = 2/(3sqrt(5)) 1/x^2 = 2(3+sqrt(5))/(9  5) 1/x^2 = (3+sqrt(5))/2 Looks like nothing, right? But let's look at ? = (1 + sqrt(5))/2 ?^2 = (1 + 5 + 2sqrt(5))/4 ?^2 = (3 + sqrt(5))/2 ?^2 = 1/x^2 x = 1/? (discarding the negative root) You can get the others by squaring and cubing this.  
May 17th, 2011, 05:40 AM  #9 
Newbie Joined: May 2011 Posts: 13 Thanks: 0  Re: Solving equation in rational roots
That's exactly what I needed! Thus, my original equation: ax1 + bx2 + cx3 = d becomes A + B*sqrt(5) = 0, where 1 and sqrt(5) form a basis. So to solve it, I set A = B = 0. Is this correct? 
May 17th, 2011, 07:22 AM  #10 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Solving equation in rational roots
Yep. (Of course you'll want to get simplified forms for x1, x2, and x3 first.)


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