1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solution of an equation: help needed

  1. Jan 27, 2007 #1
    what are the solutions of the equation [tex]\frac{1}{x} + \frac{1}{y} = \frac{1}{z}[/tex] where [tex]x, y, z \in \mathbb{Z}[/tex]

    i can see that x = y = 2z is a set of solutions. is there any other? and how do i find those?

    thanks in advance.
     
    Last edited: Jan 27, 2007
  2. jcsd
  3. Jan 27, 2007 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    What is the multiplicative inverse of both sides of the equation?
     
  4. Jan 27, 2007 #3
    i am sorry. i didn't understand how that would help. i got [tex]z = \frac{xy}{x+y}[/tex]
     
  5. Jan 27, 2007 #4
    Transform the expression as such

    [tex]\frac{x+y}{xy} = \frac{1}{z}[/tex]

    [tex]\frac{xy}{x+y} = z[/tex]


    x and y can obviously not be both odd. Also, if there's no common factor between x and y, then z is never an integer. Let's say a common factor between x and y is a. We have


    [tex]\frac{xy}{a(\frac{x}{a}+\frac{y}{a})} = z[/tex]

    In turn x/a and y/a need to have common factors for z to be an integer. If we have y > x and keep on and on, we eventually get to the point where

    [tex]a_{1}*a_{2}* a_{3} *a_{4} *a_{5}... = x[/tex]

    We conclude that y and x must have the same prime factors and that the number of times that a prime factor reoccurs in x is matched or surpassed by y. Now say y = kx, we have

    [tex]\frac{xy}{x(1+\frac{y}{x})}[/tex]

    [tex]\frac{y}{1+\frac{y}{x}}[/tex]

    [tex]\frac{kx}{1+k}[/tex]

    We can immediately see that 1 + k has to be a factor of x if z is to be an integer. Knowing that, we can construct an infinite number of pairs that satisfy the original expression; we need an y such as y = kx, and a x such that k+1 is one of its factor. For example, say k = 2, we can have x = (2+1)*7*5 and y = 2*(2+1)*7*5. So a solution is such as, if C and k are any positive integers,


    [tex] x = C(k+1), y = Ck(k+1)[/tex]
     
    Last edited: Jan 28, 2007
  6. Jan 29, 2007 #5
    sorry, but i didn't understand some of your reasonings. for example,

    why is that?
     
  7. Jan 29, 2007 #6
    Just a hunch :tongue2: I could be wrong. But the solution is valid for the case x and y have common factors. Also the kx/1+k thing is a hunch too.
     
    Last edited: Jan 29, 2007
  8. Jan 29, 2007 #7
    There is something called, "Egyptian fractions." These are all the inverse of integers. It seemed that the Ancient Egyptians did not understand fractions except in the form of 1/N, and duplicates were not used. This system was even used by the Greeks for awhile, and results in very complicated things. Here is a simple formula that is useful: 1/N = 1/(N+1) + 1/((N+1)(N)).

    Like say, we have the matter of 1= 1/3+1/3+1/3. We can expand this as: 1=1/3+1/4+1/12 +1/4+1/12 and finally: 1/3+1/4+1/12+1/5+1/20+1/13+1/156.

    This then would have satisfied the Egyptian mathematicians of the day.

    Some people believe there was SOME SENSE in this system. http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html#rhind

    For example, suppose 5 sacks of grain are to be divided up by 8 people. Well we give each person 1/2 sack and then we take the remaining sack and divide it into 8 parts. Hooray! 1/2+1/8 = 5/8.
     
    Last edited by a moderator: Apr 22, 2017
  9. Feb 6, 2007 #8
    Werg22's method in the post#4 seemed really complicated and i didn't understand half of it. can anyone please check if there is any problem with this method:

    [tex]\frac{1}{x} + \frac{1}{y} = \frac{1}{z}[/tex]

    if i put x = z+a and y = z+b, (where and b are integers) the equation becomes:

    [tex]\frac{1}{z+a} + \frac{1}{z+b} = \frac{1}{z}[/tex]

    [tex]\frac{2z + a + b}{z^2 + az + bz + ab} = \frac{1}{z}[/tex]

    [tex]2z^2 + az + bz = z^2 + az + bz + ab[/tex]

    [tex]z^2 = ab[/tex]

    now we just have to factor z2. for example if the equation was
    1/x + 1/y = 1/2007
    we would put x = 2007+a and y = 2007+b and eventually get (2007)2 = ab. and then we just have to factor (2007)2 to find a and b.
     
    Last edited: Feb 6, 2007
  10. Feb 6, 2007 #9

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    IIRC you can do a coordinate rotation of [itex] \frac {\pi} 4 [/itex] to see that this is a hyperbola. (This is just the thin lens equation with different variable names)
     
  11. Feb 7, 2007 #10

    uart

    User Avatar
    Science Advisor

    That's a good solution Murshid.

    I had another solution that I didn't post. I wont post the details but I followed a similar path to werg to get it. In particular I assummed that x and y must have a common factor (something that at the time I couldn't really justify).

    My solution was basically this. Let jk be factor of z. That is, z = ajk. Then
    x = aj (j+k)
    y = ak (j+k)
    are solutions to your equation (1/x + 1/y = 1/z).

    I like your solution a lot better. It was derived without the assumption of x and y sharing a common factor, though interestingly it can be used to prove that x and y must indeed have this common factor.

    A rough proof that x and y (as per your solution) contain a non trivial common factor is as follows.

    Let x = a + z and y = b + z,
    where a and b are any factor pair of z^2

    Since perfects squares must have all their prime factors in pairs, and since ab=z^2, then either a and b are themselves both perfect squares or they contain at least one common factor.

    Case 1 : a and b are not perfect squares.
    As above, a and b contain at least one common prime factor. Also z contains all the prime factors of z^2, so z shares the same common factors as a and b above (since z^2=ab). These common factors are thereofore also common factors of x and y.

    Case 2 : a and b are both perfect squares.
    In this case we just rewrite x and y as,
    x = sqrt(a) (sqrt(a) + sqrt(b))
    y = sqrt(b) (sqrt(a) + sqrt(b))
    So clearly x and y have a common factor.

    So the original hunch that x and y must have a common factor was indeeed correct.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...