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The question is: for any $n\geq2$, is there always a prime $p$ satifying $\varphi(n)<p\leq n$?

Here $\varphi(n)$ is the Euler totient function.

We know that there is always a prime between $n-O(n^\theta)$ and $n$, where $\theta$ can be $0.525$ (Wiki: Prime gap). Under Riemann hypothesis, one can improve this bound to $O(\sqrt n\log^2n)$. But on the other hand, there are infinite many $n$ such that $\phi(n)\geq n-C\sqrt n$ for some constant $C$ (just choose $n=p(p+k)$ where $p$ and $p+k$ are both prime; for some $k$ these $p$ are infinite). So these upper bound for prime gap don't help.

So can we prove this propsition, or give a counterexample? (or give a evidence to explan why is this hard to prove, maybe?)

(The propsition is equivalent to: if $\varphi(n)>\varphi(k)$ for all $1\leq k<n$, then $n$ is prime)

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  • $\begingroup$ By a simple program running on my laptop, there is no counterexample under $10^{10}$. $\endgroup$ – rqy Jul 8 at 16:26
  • $\begingroup$ Also, there is always a prime in the interval $[n-\sqrt{n},n]$ for all $5504 \leq n \leq 4 \cdot 10^{18}$. Hence there is no counterexample under $4\cdot 10^{18}$. $\endgroup$ – Dietrich Burde Jul 9 at 11:57
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For prime numbers $n$ the claim is trivial. If $n$ is composite then we have the upper bound $$ \phi(n)<n-\sqrt{n}. $$ If we can show that there is always a prime between $n-\sqrt{n}$ and $n$, for $n\ge 3$, then the claim follows. Unfortunately this is not yet clear, see At least one prime between N and N-(sqrtN).

Reference: Upper bound for Euler's totient function on composite numbers

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  • $\begingroup$ why "There is always a prime between $n−\sqrt{n}$ and $n$"? $\endgroup$ – rqy Jul 8 at 16:04
  • $\begingroup$ It's so hard to improve the upper bound of prime gap, I think this is not the right way (if there is a right way) $\endgroup$ – rqy Jul 8 at 16:07
  • $\begingroup$ On the other hand $\phi(n)$ can be so large for composite $n$, so it cannot be avoided, I think. $\endgroup$ – Dietrich Burde Jul 8 at 16:08
  • $\begingroup$ But maybe we can divide all natural numbers to several sorts and prove it in some steps. Not all $n-\varphi(n)$ is about $\sqrt n$. $\endgroup$ – rqy Jul 8 at 16:14

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