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Prove that if x,y, and z are integers and

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that if x,y, and z are integers and xyz=1, then x=y=z=1 or two equal -1 and the other is 1.

    2. Relevant equations


    3. The attempt at a solution
    Clearly, if I plug in 1 for each variable, or -1 in for two variables and 1 for the remaining variable, then the equation is satisfied. But how do I know these are the only solutions? This is precisely what I am trying to show. How do I sufficiently demonstrate this?

    I thought of writing the equation as ##z = \frac{1}{xy}##. But how do I show that, for any choice of integers x and y, besides 1 and -1, z will never be an integer, but a rational number?
     
  2. jcsd
  3. Oct 11, 2014 #2

    haruspex

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    What prime factors can x have?
     
  4. Oct 11, 2014 #3
    I am not sure; I imagine that it would probably be dependent upon certain circumstances, such as the present one, that xyz=1. Is that right?
     
    Last edited: Oct 11, 2014
  5. Oct 11, 2014 #4

    haruspex

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    If x has a factor p, what does that equation tell you?
     
  6. Oct 11, 2014 #5
    That you should be able to divide both sides by p....? I am not certain.
     
  7. Oct 11, 2014 #6

    HallsofIvy

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    If x> 1 then xyz> yz so yz must be less than 1.
    If x< -1 then xyz< yz so yz must be larger than -1.

    Of course, none of x, y, or z can be 0.
     
  8. Oct 11, 2014 #7
    Would there be a way to show that ##\frac{1}{xy}## is always a rational number, for integers x and y?
     
  9. Oct 11, 2014 #8

    haruspex

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    That's not exactly right since yz could be negative.
    As long as x and y are nonzero integers, it clearly is by definition. But how is that useful here? Stick with the factors hint. If a*b = c and p is a factor of a, what can you say about p and c?
     
  10. Oct 11, 2014 #9
    Well, I was thinking that this would show that there are no other integer solutions, besides the ones we already mentioned.
     
  11. Oct 11, 2014 #10

    haruspex

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    Proving a number is rational does not mean it is not an integer.
     
  12. Oct 11, 2014 #11
    Are not integers those rational numbers whose denominator is one? Well, if x and y are not one, then the denominator of 1/xy would never be one, right?
     
  13. Oct 11, 2014 #12
    To make this a little more precise, I would have to show that the only solutions to ##xy = 1## are 1 and -1. Here is my proof

    ##xy = 1##

    ##x^2 y = x##

    ##x^2 y - x = 0##

    ##x(y-1) = 0##

    This implies that x=0 or y-1 = 0. But x cannot be 0, as this would not satisfy the equation. Hence, it must be true that ##y-1 = 0 \implies y=1##. If y=1, then ##x \cdot 1 = 1 \implies x =1##. Notice, however, that I do not get the negative one solution. What is wrong with my proof?
     
  14. Oct 11, 2014 #13

    vela

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    ##x^2 y - x \ne x(y-1)##
     
  15. Oct 11, 2014 #14
    Ah, yes. I see. So, proving xy=1 is, I imagine, similar to proving that xyz=1. Now, supposing I could prove that xy= 1 implies x=y=1 or x=y=-1, would what I said in this post

    still be valid?
     
  16. Oct 11, 2014 #15

    haruspex

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    Yes, if you add that step it works, but it is rather a long way round. Do try to answer my question about factors.
     
  17. Oct 11, 2014 #16
    Would it be that p is a factor of c?
     
  18. Oct 11, 2014 #17

    haruspex

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    Yes. So if c=1 then...?
     
  19. Oct 11, 2014 #18
    p would have to be one.
     
  20. Oct 12, 2014 #19

    haruspex

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    If positive, yes, but it could also be...?
     
  21. Oct 12, 2014 #20
    Or negative one.
     
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