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    What a title.

    This is a small part of my solution to a larger question, if I can prove this then I prove the rest, and my brain has shut down. This is probably extremely easy.

    Suppose x^a, x^b are integral for two integers a,b and gcd(a,b) = 1. Is x an integer?
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    No. Take a=b=-1 and x = 1/2.
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    Sorry a,b are positive coprime integers.
    And another mistake, x^a and x^b are integers but not necessarily equal.
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    Yes.

    x^a =A,\;x^b =B\Rightarrow A^b=B^a\;\;(\star)

    Write A=p_1^{\alpha_1}\cdots p_i^{\alpha_i} and B=q_1^{\beta_1}\cdots q_j^{\beta_j} with p_k,\;q_k primes.

    By (\star):\;i=j,\;p_k=q_k. This now implies b\alpha_k=a\beta_k for all k. Since \gcd(a,b)=1 we must have \alpha_k=an_k and \beta_k=bn_k

    Hence A=(p_1^{n_1}\cdots p_i^{n_i})^a and B=(p_1^{n_1}\cdots p_i^{n_i})^b

    In other words, x=p_1^{n_1}\cdots p_i^{n_i}
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    If (a,b) = 1 then there exist m,n with am + bn = 1. So x^{am + bn} = x = A^mB^n. So x is rational, and thus by a special case of the rational root theorem it must be an integer.
 
 
 
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