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Irrational equations

  1. Feb 10, 2009 #1
    Show that the equation 4x^(2) + 5y^(2) = 2 has no rational solutions.

    Can this be done graphically?
     
  2. jcsd
  3. Feb 10, 2009 #2

    Hurkyl

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    If you have a rigorous graphical method, then of course. Though, I'm hard pressed to think of any that would be applicable....
     
  4. Feb 10, 2009 #3

    MathematicalPhysicist

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    Suppose the equation has rational roots:
    4x^(2) + 5y^(2) = 2
    x=p1/q1
    y=p2/q2
    gcd(p1,q1)=gcd(p2,q2)=1
    4(p1^2/q1^2)+5(p2/q2)^2=2
    [4(p1q2)^2+5(p2q1)^2]/(q2q1)^2=2
    4(p1q2)^2+5(p2q1)^2=2(q2q1)^2
    then the question becomes does the equation:
    4a+5b=2c
    where a,b,c are natural numbers has solution.
    4(a+b)+b=2c
    because 2c is even b must be even also, but then b=(4a-2c)/5 then 4a-2c is divisble by 5, i.e 4a-2c=0 or a and c are multiples of 5.
    if b=0 then c=2a i.e (q2q1)^2=2(p1q2)^2 but this isn't possible cause sqrt(2) is irrational.
    Now you need to check the option of multiples of 5 and get a contradiction, perhaps someone else with more intel in number theory will chip in with this question.
     
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