Suppose I show that: $$x^{f(z)/g(z)} = y \pmod{4}$$ is impossible for some given positive integers $x$ and $y$, where, \begin{align*} f(z) &= \phi(4) k_1(z) + 1 \\ &= 2 k_1(z) + 1\\ g(z) &= \phi(4) k_2(z) + 1 \\ &= 2 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 $y$ ?
I think the answer is negative, you can't derive $x^d \ne y $ from $x^{f(n)/g(n)}\not \equiv y \pmod{4} $ for every $n$ and $$\lim_{n\to \infty} \frac{f(n)}{g(n)} \to d $$ Look at this example $$ x= 2, y = 3, d = \frac{\log 3}{\log 2} $$ as $d$ is irrational (see Hardy and Wright p. 162), we can find rational numbers $f(n)/g(n)$ with odd numerator and denominator that converge to $d$, (for example if $n$ is odd and $$ \frac{a_n}{n} < \frac{\log 3}{\log 2} < \frac{a_n+1}{n} $$ then pick $g(n) = n$ and $f(n) = a_n$ if $a$ is odd or $f(n)=a_n+1$ otherwise).
We have obviously $$ 2^{f(n)} \not\equiv 3^{g(n)} \pmod{4} $$ because the left hand side is even and the right hand side is odd however $$ 2^d = 3$$