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Is there an analogue to Bertrand's Postulate for primes for perfect powers?

Bertrand's Postulate: $\forall x \gt 1 \in \mathbb{Z}, \exists p \in \mathbb{P}$ such that $x < p < 2x$.

For perfect powers, could we say something like $\exists k,m,\lambda \in \mathbb{Z}, m \gt 2$ such that $\forall x \in \mathbb{Z}, x \gt 2$, $x \lt k^m \le \lambda x$? Obviously if $\lambda \ge x^2$, we have $(k,m) = (x, 3)$ as a satisficatory perfect power in the range.

This question came up while thinking about this other MSE question I had posted and received an answer. I am interested in knowing how close to $x$ are we likely to find a perfect power $k^m \gt x$ with an exponent $m\gt2$?

vvg
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    Your quantifiers are jumbled. Clearly there aren't $k,m,\lambda$ that work for all $x$. I suppose you meant $\lambda$ to be independent of $x$ (else, as you remark, the problem isn't interesting), but $k$ (and presumably $m$ as well) should depend on $x$. – lulu Oct 16 '22 at 13:49
  • Most of the perfect powers are perfect squares. and they satisty most properties primes also do , for example for every $\epsilon>0$ , there is a positive integer $n_0$ such that there is a perfect square strictly between $n_0$ and $n_0(1+\epsilon)$ – Peter Oct 16 '22 at 16:22
  • @lulu, yes. I intended $\lambda$ to be independent of $x$ analagous to the $x < p < 2x$ in Bertrand's postulate. I just used the $\lambda = x^2$ to show an exponent $m > 2$. As Peter says, squares dominate. Question is how close are higher powers $m > 2$ from any chosen $x$ – vvg Oct 16 '22 at 18:23

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