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B Ab = a + b

  1. Dec 22, 2016 #1
    There is the equation

    ab = a + b (the solutions have to be integers)

    A solution to this is a= b= 2.

    What are other solutions? Are there other solutions?
     
  2. jcsd
  3. Dec 22, 2016 #2

    Orodruin

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    Why don't you solve for b in terms of a and think about what would be necessary for b to be an integer if a is?
     
  4. Dec 22, 2016 #3

    blue_leaf77

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    One more.
     
  5. Dec 22, 2016 #4
    it would be that

    b= a / (a - 1)
    thus b has to be a multiple of a - 1

    I assume that the other solution is a=b=0
     
  6. Dec 22, 2016 #5

    Orodruin

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    No, it is ##a## that needs to be a multiple of ##a-1## in order for ##b## to be an integer!
     
  7. Dec 22, 2016 #6
    what pairs of integers satisfy that? I can't for the life of me think of any!
     
  8. Dec 22, 2016 #7

    jedishrfu

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    Did you try to simply insert common integers into the equation?

    You got one solution of a=b=2 from a^2=(a+a) which can be converted to a quadratic expression to factor right?
     
  9. Dec 22, 2016 #8

    fresh_42

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    So prove it.
    ##a## being a multilple of ##a-1## means ##a=b\cdot (a-1)## and since ##a## is the greater number, ##b## has to be positive.
    Can you show why ##b## cannot be greater than ##2##?
     
  10. Dec 22, 2016 #9
    if b is greater than 2, say 3, then a= 3a - 3. no integer satisfies this from 1 to 10 and has a gets bigger the 3 is a smaller proportion of a so it can be assumed that this is the case
     
  11. Dec 22, 2016 #10

    fresh_42

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    Well, .... yes. But the restriction to numbers until ##10## lacks a bit of rigor. ##a=b(a-1) > 2(a-1) = 2a -2## and thus ##a<2## would be more general and also leaves you with the cases ##a \in \{0,1\}\, ##or## \, b \in \{0,1,2\}## which you can handle manually. Or you proceed along the lines @jedishrfu has pointed out in #7.
     
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