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I asked a similar question to this recently. Here, I consider an arbitrary, but fixed, modulus m, which is relatively prime to x and y. Can anybody extend the answer given in the previous question?

So, my question is the following. Suppose I show that: $$x^{f(z)/g(z)} = y \pmod{m}$$ is impossible for some given positive integers $x$ and $y$, and integer $m>4$, where m is relatively prime to both x and y, and $f(z)$ and $g(z)$ are defined by, \begin{align*} f(z) &= \phi(m) k_1(z) + 1 \\ g(z) &= \phi(m) k_2(z) + 1 \\ \end{align*} and $k_1(z)$ and $k_2(z)$ are integer functions, that approach infinity, such that $f(z)/g(z)$ approaches some irrational number. Can I then say, the equation: $$x^{f(z)/g(z)} = y$$ has no solutions integer solutions, with the same $x$ and $y$, as $z$ goes to infinity as well?

That is, if I let, $$d = \lim_{z->\infty}\frac{f(z)}{g(z)}$$ be the irrational number in the limit, then would it be true that, $$x^d \neq y$$ for the same $x$ and $y$ ?

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@Arturo: Thanks for adding that link for me! I am learning the notation much more rapidly by example than I would have by reading all that documentation. –  user7105 Feb 16 '11 at 4:46
    
No problem; as you can see from the Copy Editor badge, of which I am currently the only proud owner, I do edits a lot. –  Arturo Magidin Feb 16 '11 at 4:47
    
My thoughts on this are that because I imposed the additional restriction that $m$ be relatively prime to $x$ and $y$, the limit will actually work out and be equivalent to the equation. Because, I think, but I am not entirely sure, that the lack of solution to the congruence implies a lack of solution to the integer equation, if $m$ is relatively prime. That is the only step I am unsure of. –  user7105 Feb 16 '11 at 4:56
    
It is somewhat imprecise to say "if $m$ is relatively prime"; 'relatively prime' is relation between numbers, so you should really really say 'relatively prime' to what (I think, "to $xy$"). –  Arturo Magidin Feb 16 '11 at 5:05

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