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Proof about a diophantine equation

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that the Diophantine equation
    [itex] 3y^2-1=x^2[/itex]
    has no solution.
    3. The attempt at a solution
    If we factor the left side as
    [itex] (\sqrt{3}y+1)(\sqrt{3}y-1) [/itex]
    I was going to try to argue that x could only divide one of those.
    But I dont know if thats a good place to start.
     
  2. jcsd
  3. Feb 9, 2013 #2

    Dick

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    No. It's a really awful place to start. Diophantine means you are supposed to get solutions that are integers. You know that, don't you? What other ways have you seen to prove stuff like this?
     
    Last edited: Feb 9, 2013
  4. Feb 9, 2013 #3
    I know they need to be integers, I was trying to think of a way to argue that they had different amount of prime divisors. Or maybe it has something to to with their gcd's.
     
  5. Feb 9, 2013 #4

    Dick

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    Those are more reasonable things to think about. How about trying to see if you can solve the equation modulo some number?
     
  6. Feb 9, 2013 #5
    What about this if x is even then [itex] 3y^2 [/itex]must be odd
    so we have [itex]3y^2=2a+1[/itex] and then we
    have [itex] 3y^2-1=2a=x^2 [/itex]
    if x=2 then this would work. but if a has more factors than just 2.
    if a is a perfect square then it will either contain a factor of 2 or not in either case
    it will have an odd number of factors of 2 but x has an even number of factors.
    So this wont work.
     
  7. Feb 9, 2013 #6

    Dick

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    Sure. If a is a perfect square then 2a=x^2 can't be solved. That's basically how to prove sqrt(2) is irrational. But why would you think a needs to be a perfect square? Here's a big hint. If a is any integer, what can you say about a^2 modulo 4?
     
  8. Feb 11, 2013 #7

    CAF123

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    Can we not just consider the equation ##x^2 \equiv -1 ##mod ##3##, and show from here that this has no solutions?
     
  9. Feb 11, 2013 #8

    Dick

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    You could also do that. Sure.
     
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