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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 = ((3-sqrt(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 anti-clockwise. 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 = (n-2)*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)*(n-2) 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((3-sqrt(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
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Re: Solving equation in rational roots

Quote:
 Originally Posted by rhseiki I wonder whether we can assume that the powers of cos(72)( are linearly independent in the ring of rational numbers
(I assume you mean the cosine of 72 degrees.)

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((3-sqrt(5))/2) be written in the form: a + b*sqrt(5), where a and b are rationals. Or cos(72)^3 = ((3-sqrt(72))/2)*sqrt((3-sqrt(5))/2). May 16th, 2011, 09:23 PM   #6
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Re: Solving equation in rational roots

Hello, rhseiki!

Quote:
 [color=beige]. . [/color] 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 non-trivial 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
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Re: Solving equation in rational roots

Quote:
 Originally Posted by rhseiki x1 = ((3-sqrt(5))/2)^1/2 x2 = x1^2 x3 = x1^3
Writing x for x1,

x = ((3-sqrt(5))/2)^1/2
1/x = (2/(3-sqrt(5)))^1/2
1/x^2 = 2/(3-sqrt(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.) Tags equation, rational, roots, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post edwinandrew Algebra 1 November 27th, 2013 04:28 PM boomer029 Algebra 3 March 18th, 2012 10:20 PM Pumpkin99 Algebra 2 August 11th, 2011 01:29 PM RKJCHENNAI Abstract Algebra 5 March 5th, 2010 04:12 PM DinkyDoeDoe Abstract Algebra 1 June 30th, 2009 11:08 PM

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