We start with the following observation: if $ a^x = b^y$ with $ a, b, x, y$ positive integers and $ (x, y) = 1$ ten $ a = c^y$ for some $ c$. This is almost obvious: take prime $ p|a$ and $ p^{\alpha}||a$. Then $ y|\alpha x$, so $ y|\alpha$. We can write $ r = \frac {a}{b}$ with $ a, b > 0$ and $ (a,b) = 1$. Suppose that $ \sqrt [r - 1]{r} = \frac {c}{d}$ for some positive, relatively prime integers $ c, d$. Then we of course have $ \frac {a^b}{b^b} = \frac {c^{a - b}}{d^{a - b}}$ which clearly forces $ a^b = c^{a - b}$ and $ b^b = d^{a - b}$. We must have $ a > b$ and since $ (a, b) = 1$ we also have $ (b, a - b) = 1$, so it follows from our observation that $ a = x^{a - b}$ and $ b = y^{a - b}$ for some positive integers $ x, y$ with $ x > y$. But if $ a - b > 1$: $ a - b = x^{a - b} - y^{a - b} \geq (y + 1)^{a - b} - y^{a - b} = y^{a - b - 1}(a - b) + ... + 1 > a - b$ Hence $ a - b = 1$ and it's also easy to check that it is the sufficient condition. So: $ r=\frac{p+1}{p}$ with $ p$ positive integer.

??? The order of a root is a NATURAL number, not a fraction...

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=376440& think this is from 1999 rather than 2000...

The final result is correct.

Peter wrote: Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational. Do you mean that r-1 can be a rational number that isn't an integer number?

Peter wrote: Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational. Rewrite $r^{\frac{1}{r-1}}$. For Gelfond's theorem $ a^b $ with $a$ rational can be rational only if b is an integer or if $ a=c^ \frac{1}{b} $ for some rational c. Under the conditions of the problem, b is an integer only if r=2 or r=1+1/n, n natural number.

ctumeo wrote: Peter wrote: Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational. Rewrite $r^{\frac{1}{r-1}}$. For Gelfond's theorem $ a^b $ with $a$ rational can be rational only if b is an integer or if $ a=c^ \frac{1}{b} $ for some rational c. But this is clearly impossible under the conditions of the problem. If you don't agree, post a counterexample (i.e. any rational values for r). What about $r=2$?

Weeks_behind wrote: ctumeo wrote: Peter wrote: Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational. Rewrite $r^{\frac{1}{r-1}}$. For Gelfond's theorem $ a^b $ with $a$ rational can be rational only if b is an integer or if $ a=c^ \frac{1}{b} $ for some rational c. But this is clearly impossible under the conditions of the problem. If you don't agree, post a counterexample (i.e. any rational values for r). What about $r=2$? I've edited my post.