Guest Posted March 2, 2010 Report Share Posted March 2, 2010 (edited) Each of X,Y and Z is a positive integer satisfying: X2+Y2+1 = X*Y*Z. Determine all possible value(s) that Z can assume. Edited March 2, 2010 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 jazzship Posted March 2, 2010 Report Share Posted March 2, 2010 3 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 2, 2010 Report Share Posted March 2, 2010 (edited) x=2 y=5 z=3 Each of X,Y and Z is a positive integer satisfying: X2+Y2+1 = X*Y*Z. Determine all possible value(s) that Z can assume. One possible solution... Edited March 2, 2010 by rover Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 2, 2010 Report Share Posted March 2, 2010 (edited) for example, subbing x =2 5 + y^2 = z*2y y^2 - 2zy + 5 = 0 (y - 1)(y - 5) = 0 => -2zy = -6y z = 3 similarly, for x=1: y^2 -zy + 2 = 0 (y-1)(y-2) = 0 => z=2 we can see the general solution is such that z will equal any positive integer that fits: [sum of 2 factors of (x^2+1) ] / x, where x is a positive integer while for x=1,2 this leads us to solutions for z by factoring (x^2 +1) into {1, (x^2+1)}, for values of x >= 3, there will be no such solution possible. That still leaves the possibility that the sum of 2 other factors (in between) will be divisible by x. However, that is also not possible unless z = 3. Still working on trying to prove this. for example: x = 13: y^2 - 13zy +170 = 0 (y-34)(y-5) = 0 => 13z = 39 z = 3 the other factor pairs, don't produce a solution (1,170), (10, 17) , (2, 85) Edited March 2, 2010 by jlf278 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 2, 2010 Report Share Posted March 2, 2010 for example, subbing x =2 5 + y^2 = z*2y y^2 - 2zy + 5 = 0 (y - 1)(y - 5) = 0 => -2zy = -6y z = 3 similarly, for x=1: y^2 -zy + 2 = 0 (y-1)(y-2) = 0 => z=2 we can see the general solution is such that z will equal any positive integer that fits: [sum of 2 factors of (x^2+1) ] / x, where x is a positive integer while for x=1,2 this leads us to solutions for z by factoring (x^2 +1) into {1, (x^2+1)}, for values of x >= 3, there will be no such solution possible. That still leaves the possibility that the sum of 2 other factors (in between) will be divisible by x. However, that is also not possible unless z = 3. Still working on trying to prove this. for example: x = 13: y^2 - 13zy +170 = 0 (y-34)(y-5) = 0 => 13z = 39 z = 3 the other factor pairs, don't produce a solution (1,170), (10, 17) , (2, 85) re-check your work for x=1, you'll find that z does in fact equal 3... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 2, 2010 Report Share Posted March 2, 2010 if the equation is rearranged to (x^2+y^2+1)/(x*y)=z, then z can assume an infinite number of values based on what x and y are. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 2, 2010 Report Share Posted March 2, 2010 if the equation is rearranged to (x^2+y^2+1)/(x*y)=z, then z can assume an infinite number of values based on what x and y are. Your conclusion does not support the original requirement of the problem which is X, Y, and Z be positive intergers. So if x=1, y=4 gives (1 + 16 + 1)/4 = 18/4 = 4.5 which means z is not an interger. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 3, 2010 Report Share Posted March 3, 2010 Your conclusion does not support the original requirement of the problem which is X, Y, and Z be positive intergers. So if x=1, y=4 gives (1 + 16 + 1)/4 = 18/4 = 4.5 which means z is not an interger. Not only that, but Z can never be between -2 and 3 for any real values of x and y. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 10, 2010 Report Share Posted March 10, 2010 solution.pdf Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 10, 2010 Report Share Posted March 10, 2010 solution.pdf You're proof is more complicated than necessary. One need only consider the three cases: 1) x=1,y=1 => z=3 2) x=1,y>1 => y=2,z=3 3) x>1,y>1 => z=DNE Thus, z=3. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 11, 2010 Report Share Posted March 11, 2010 You're proof is more complicated than necessary. One need only consider the three cases: 1) x=1,y=1 => z=3 2) x=1,y>1 => y=2,z=3 3) x>1,y>1 => z=DNE Thus, z=3. You're proof is more complicated than necessary. One need only consider the three cases: 1) x=1,y=1 => z=3 2) x=1,y>1 => y=2,z=3 3) x>1,y>1 => z=DNE Thus, z=3. Hi vinays84, thanks for commenting on my post but, I'm sorry, I have to disagree with your solution. What do you mean by DNE? In principle, I can imagine only two reasonable answers for that question: either DNE="Does Not Extist" or DNE = 3. However, as I pointed out in my solution, the equation holds for (x, y, z) = (2, 5, 3). Hence, it's wrong to say "3) x>1,y>1 => z Does Not Exist" Therefore, the only possibility remaining is "3) x>1,y>1 => z=3" Then I ask how do you prove that? Essentially, this implication is the original problem or at least a main part of it, since the case x<1 or y<1 is trivial (as you have noticed in 1 and 2). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 12, 2010 Report Share Posted March 12, 2010 For the 2 hours I was expanding, simplifying, dividing, multiplying and all kind of stuff with x2 + y2 + 1 = x*y*z. I somehow found that xy-1 - 3 = z2. From here: z2 must be greater or equal to 0 (To be a real number). From the original equation we find, that z must be greater or equal to 3, so for any z greater or equal to 3 there is a xy-1 - 3, equal to z2. Quote Link to comment Share on other sites More sharing options...
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Each of X,Y and Z is a positive integer satisfying: X2+Y2+1 = X*Y*Z.
Determine all possible value(s) that Z can assume.
Edited by K SenguptaLink to comment
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