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    find all integers a, b and c that satisfy  a(2^{1/2})-b=c(3^{1/2})
    I've factorised it so that  3^{1/2}(a/(1.5^{1/2})-b/(3^{1/2})) is on the left side. Where do I go from here? (assuming that I'm right)
    Sorry for the poor latex, first time

    What level is this question? C1?
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    (Original post by Daiblain)
    find all integers a, b and c that satisfy  a(2^{1/2})-b=c(3^{1/2})
    I've factorised it so that  3^{1/2}(a/(1.5^{1/2})-b/(3^{1/2})) is on the left side. Where do I go from here? (assuming that I'm right)
    Sorry for the poor latex, first time

    What level is this question? C1?
    You're over complicating it.
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    (Original post by Daiblain)
    find all integers a, b and c that satisfy  a(2^{1/2})-b=c(3^{1/2})
    I've factorised it so that  3^{1/2}(a/(1.5^{1/2})-b/(3^{1/2})) is on the left side. Where do I go from here? (assuming that I'm right)
    Sorry for the poor latex, first time

    What level is this question? C1?
    Think about it. What do you expect the solution to be? With non-routine problems it's important to think about the problem for a bit before rushing into blind algebraic manipulations.
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    Since it's asking for 'all' the integers, would the answer be an inequality? I have no clue
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    (Original post by Daiblain)
    Since it's asking for 'all' the integers, would the answer be an inequality? I have no clue
    No, the only way you can add an integer multiple of the square root of two to an integer and expect to get an integer multiple of the square root of 3 is if a,b,c=0
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    (Original post by Zacken)
    No, the only way you can add an integer multiple of the square root of two to an integer and expect to get an integer multiple of the square root of 3 is if a,b,c=0
    So there are no solutions other than 0 for all of them? And the only reason you can't add them together is because both sides are irrational?
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    (Original post by Daiblain)
    So there are no solutions other than 0 for all of them? And the only reason you can't add them together is because both sides are irrational?
    Yep. And no, it's because of a more technical reason: namely them being linearly independent.
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    (Original post by Zacken)
    Yep. And no, it's because of a more technical reason: namely them being linearly independent.
    Thanks alot!
 
 
 
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