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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

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    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: Jan 29, 2007
  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

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    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

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    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.
     
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