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 October 26th, 2007, 11:30 AM #1 Newbie   Joined: Oct 2007 Posts: 10 Thanks: 0 Three problems to solve Hi everybody, This is what children in Bulgaria from the 10th grade have to be able to solve if they wish to take part in a national competition in mathematics. Unfortunatelly I wasn't able to solve these three tasks: 1. Prove that this number: 11..1(2^n times 1) can be divided by at least 2^n different numbers. 2. Say which numbers are answers to each of the three equations: x+y.y=z.z.z x.x+y.y.y=z.z.z.z x.x.x+y.y.y.y=z.z.z.z.z 3.Solve the equation where a is a parameter: (x*x-a)(x*x-a)-6x*x+4x+2a=0 I hope they are not so difficult for you! And I hope you will help me soving them!
October 26th, 2007, 12:24 PM   #2
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Re: Three problems to solve

Quote:
 Originally Posted by Rooonaldinho 2. Say which numbers are answers to each of the three equations: x+y.y=z.z.z x.x+y.y.y=z.z.z.z x.x.x+y.y.y.y=z.z.z.z.z
Are you forgetting some constraints? (0, 0, 0), (1, 0, 1), and (0, 1, 1) are all obvious solutions.

Edit: Are we supposed to characterize all solutions of the equations? Are we supposed to solve them individually or as a system?

 October 26th, 2007, 01:01 PM #3 Newbie   Joined: Oct 2007 Posts: 10 Thanks: 0 Unfortunatelly, I'm not forgetting anything. Yes, it's a system. I'm sorry I haven't said it is. As for the obvious solutions (0,0,0),(1,0,1) and (0,1,1) I know about them. But I would like to know if there are any other solutions and how to get to them. If there aren't any other tell me how do you understand there aren't.
 October 26th, 2007, 01:29 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 Well it's pretty easy to check that the only solutions with x, y, or z being zero are (0, 0, 0), (0, 1, 1), (1, 0, 1), and (-1, -1, 0). You should verify this. So otherwise assuming that none of the three are zero, you can combine the first and third equations and compare to the square of the middle, removing the z term. This reduces to 2xy = x^2 + y^2, which gives you a very useful (three-character) equality if you think about it. Combining this with any two of the remaining equations gives you another (three character) equality. This reduces the solution to solving a quadratic and checking. I get six solutions overall.
October 26th, 2007, 01:36 PM   #5
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Re: Three problems to solve

Quote:
 Originally Posted by Rooonaldinho 3.Solve the equation where a is a parameter: (x*x-a)(x*x-a)-6x*x+4x+2a=0
Likewise, I don't understand this problem. What are we to solve for, and under what conditions? Closed form solution for x in a? Exhaustive list of solutions for a, given that the relation holds for all x? Closed form solution for a in x?

 October 26th, 2007, 02:58 PM #6 Newbie   Joined: Oct 2007 Posts: 10 Thanks: 0 Well the first one, you should explore the equation for each value of a and say what value(s) of x will be a solution or say that there's no solution for the current value of a. And because the values of a are countless, the answer will be for example:" for a>-10 there are no solutions, for a=3 there are two solutions x1=3 , x2=-3 and for a>3 again no solution". Someting like that. But for the second task, I didn't understand how you got to 2xy = x^2 + y^2, I tried to divide and multiply the the first and the third equation, if that is to combine them but nothing happened. Will you please write all of the lines of your solution to follow it more carefully and understand what exactly is your solution.
 October 26th, 2007, 03:50 PM #7 Newbie   Joined: Oct 2007 Posts: 10 Thanks: 0 Okay, I got it how I can get to 2xy=y^2+x^2. But it means that x=y because the it transforms like this: 0=y^2-2xy+x^2 and then 0=(y-x)^2 so x=y right? Then the three equations are 1 x+x.x=z.z.z 2 x.x+x.x.x=z.z.z.z 3 x.x.x+x.x.x.x=z.z.z.z.z
October 26th, 2007, 07:18 PM   #8
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Quote:
 Originally Posted by Rooonaldinho Will you please write all of the lines of your solution to follow it more carefully and understand what exactly is your solution.
No, I'd hate to rob you of the ability to solve it yourself. I have just enough to help you along.

Quote:
 Originally Posted by Rooonaldinho Okay, I got it how I can get to 2xy=y^2+x^2. But it means that x=y because the it transforms like this: 0=y^2-2xy+x^2 and then 0=(y-x)^2 so x=y right?
Good. I actually got to that by noting that x and y were of the same sign (both positive or both negative) and that xy >= x^2 and xy >= y^2, but your way is probably better -- or at least more standard.

The next step is to manipulate those equations. You can divide if you like, since I already showed all the cases where any of the three variables are zero. See any kind of pattern, anything you can do to reduce it to one or two equations?

 October 27th, 2007, 10:03 AM #9 Newbie   Joined: Oct 2007 Posts: 10 Thanks: 0 I think I solved the second one I think I solved the second one. It turned out that x is even equal to z. The answers are (0;0;0),(0;1;1),(1;0;1),(-1;-1;0),((1-sqrt(5))/2;(1-sqrt(5))/2;(1-sqrt(5))/2))) and ((1+sqrt(5))/2;(1+sqrt(5))/2;(1+sqrt(5))/2))). Tell me if that's right and thank you for the help! Can you also give me an idea about the other tasks too?
October 27th, 2007, 05:42 PM   #10
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Re: I think I solved the second one

Quote:
 Originally Posted by Rooonaldinho Tell me if that's right and thank you for the help!
That looks right. I remember getting a similar answer (in terms of phi,the golden ratio, for the last two).

How far can you simplify #3?

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