<s>[INST] show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. I am trying to show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. The reverse implication is easy, but the forward implication is tricky. It seems essentially to state the the (n-1)th root of a natural number n is not an integer for n>2.
 [/INST] Define $f(x) = \dfrac{\ln x}{x-1}$. Then, $f'(x) = -\dfrac{1}{(x-1)^2}\ln x + \dfrac{1}{x-1} \cdot \dfrac{1}{x} = -\dfrac{x(\ln x - 1)+1}{x(x-1)^2}$. 
For $x \ge 3$, we have $f'(x) < 0$, so $f$ is decreasing on $[3,\infty)$. 
Hence, for all $c \ge 3$, we have $f(c) \le f(3)$, i.e. $\dfrac{\ln c}{c-1} \le \dfrac{\ln 3}{2}$. 
Exponentiating both sides to get $c^{\tfrac{1}{c-1}} \le \sqrt{3} < 2$. 
Trivially, $c^{\tfrac{1}{c-1}} > 1$ for integers $c \ge 3$. Therefore, $1 < c^{\tfrac{1}{c-1}} < 2$ for all integers $c \ge 3$. 
Hence, $c^{\tfrac{1}{c-1}}$ is not an integer for any integer $c \ge 3$.
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