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!

Homework Help: Integer solution

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data

    let [tex]x,y \in N[/tex]. Solve the equation [tex]\frac{1}{x}+\frac{1}{y}=1[/tex]

    2. Relevant equations


    3. The attempt at a solution

    so i can see x=y=2 is the solution, hmm, isn't there any other solution.

    and, this is one of three of my assignment question for 15% continuous assessment, does it seem too easy? should i prove that they are no other solution other than x=y=2?? help
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2
    You shouldn't prove that there is no other solution because there are other solutions. How about x = 3 and y = (3/2)? or x = 6 and y = (5/6) ??? This is an implicit function describing a curve (in this case a straight line) of some sort, which means there are infinite points (values of x and y) that satisfy the equation. Try making it an explicit function by solving for y (or x if you want) and then see what you get.
  4. Sep 7, 2010 #3
    owh i forgot to wrote it, it's on the title "integer solution" hmm lemme edit
  5. Sep 7, 2010 #4
    Hint: If x > 2, then [itex]\frac{1}{x}[/itex] is ... .

  6. Sep 7, 2010 #5
    If [tex]x>2[/tex] then [tex]\frac{1}{x}+\frac{1}{y}<\frac{1}{2}+\frac{1}{y}[/tex] and if [tex]y \geq 2[/tex], then [tex]\frac{1}{x}+\frac{1}{y}<\frac{1}{2}+\frac{1}{2}=1[/tex]

    similarly if [tex]y>2[/tex] and [tex]x \geq 2[/tex], so either case is not a solution. should i do like that?
  7. Sep 7, 2010 #6
    Yes, that's the right idea.

  8. Sep 7, 2010 #7
    Petek illustrates an important method, but this diophantine equation can be solved explicitly. Simply clear denominators and rearrange to get xy - x - y = 0. Now try to add a constant to both sides so that you can factor the left hand side as a product of two linear factors, similar to completing the square.
  9. Sep 8, 2010 #8
    cool, i get

    [tex](x-1)(y-1)=1[/tex] implies [tex](x-1)=1 \ and\ (y-1)=1[/tex]

    but what if

    [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1[/tex], i can't solve it explicitly can't i?

    so i know[tex]x=y=z=3[/tex] is a solution

    and i need to verify, [tex]x \in N-(3)\ ,y \in N-(3)\ ,z \in N-(3)[/tex] is not a solution

    so for [tex]x=1,2[/tex] easy to verify

    if [tex]x>3 \ \ \ \ [/tex] [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{3}+\frac{1}{y}+\frac{1}{z}[/tex] so when y=1,z=2 and y=2,z=1 and y=1,z>3 and y=2,z>3 and y>3,z=1 and y>3,z=2, is not a solution(i have to verify all these cases?)

    and then repeat this tedious cases for y>3 and z>3

    is that the way?
  10. Sep 8, 2010 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Actually, 1 factors in two different ways over the integers:
    1 = 1*1
    1 = -1 * -1
  11. Sep 8, 2010 #10
    owh yes, thank ;P

    so (x-1)=-1 =>x=0 similarly y=0, which is not a solution.

    but how about this

    is this argument correct? T_T
  12. Sep 8, 2010 #11


    User Avatar
    Homework Helper

    As y is an integer, we know that the above fraction is an interger, so x-1 must perfectly divide x, when can this happen?
  13. Sep 8, 2010 #12
    hmm since gcd(x-1,x)=1 so x/(x-1) to be integer means x-1 must equal to 1 or -1, right?

    hey, if i solve explicitly like this or like

    that's the only solution right? and other integer is not a solution?? and i don't need to verify others is not a solution right?? right right??? is it right?
  14. Sep 8, 2010 #13


    User Avatar
    Homework Helper

    Correct, but you can get rid of one of the options with a bit of thought. I was mainly thinking that if x is an integer then the greatest divisor must be x/2 if x is even or less if x is odd.
  15. Sep 8, 2010 #14
    wait what you mean by

    and thanks for this clue
    i'm interesting to prove that statement ;P
    Last edited: Sep 8, 2010
  16. Sep 8, 2010 #15


    User Avatar
    Homework Helper

    I always loathed number theory. Look at the what x-1=1 and x-1=-1 and see what it means with the original equation.

    As with the rule, it was just an observation. I have no idea how to prove it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook