4
$\begingroup$

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 $y$ ?

  • 0
    No problem, other than the fact that we're back at square one, right?2011-02-15

1 Answers 1

3

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$

  • 0
    I really wonder about mod$p$for some prime p.2012-11-12