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 $y$ ?

  • $\begingroup$ Thanks to whoever formatted my equations. I'm still trying to figure out math notations. $\endgroup$ – user7105 Feb 15 '11 at 5:41
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    $\begingroup$ That $x^d=y$ has no solutions with $x$ an integer, $x\neq 0,1$, $d$ irrational, and $y$ algebraic (in particular, for $y$ integer) follows from the Gelfond-Schneider theorem. $\endgroup$ – Arturo Magidin Feb 15 '11 at 5:42
  • $\begingroup$ Doesn't the Gelfond-Schneider theorem refer to x^y != d by your definitions, rather than x^d = y? I mean, if we assume y is relatively prime to x, certainly d would have to be irrational, right? $\endgroup$ – user7105 Feb 15 '11 at 5:58
  • $\begingroup$ Sorry; you'd need $d$ to be algebraic, not just irrational, to apply Gelfond-Schneider. My mistake. $\endgroup$ – Arturo Magidin Feb 15 '11 at 6:06
  • $\begingroup$ No problem, other than the fact that we're back at square one, right? $\endgroup$ – user7105 Feb 15 '11 at 6:09

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 $$ 2^d = 3$$

  • $\begingroup$ You are absolutely right. I see it now. Thank you! On a closely related note, do you know if there is any way to express the limit in the modulus, such that it does correspond to the equation? $\endgroup$ – user7105 Feb 16 '11 at 2:42
  • $\begingroup$ To define limits in modular equations you need a topology, I think that what you are lookimg is $p$-adic analysis. But even with a proposition like $\lim x^{f(n)/g(N)} \ne y$ in $\mathbb{Q}_p$ (if that can be properly defined) for every $p$ I'm not sure if you could derive that $x^d \ne y$ in $\mathbb{R}$ $\endgroup$ – Esteban Crespi Feb 16 '11 at 7:27
  • $\begingroup$ I really wonder about mod p for some prime p. $\endgroup$ – mick Nov 12 '12 at 17:28

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