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Does there exist $N >2024 $ such that $\phi (n) > \phi (N) , \space \forall n>N$ ? Here $n , N \in \mathbb{N} $ and $\phi$ is the Euler's Totient Function.

I think such a $N$ doesn't exist due to the non-monotonicity of $\phi (n)$.

But if it is false , I have to produce a function $f(N):=n$ such that $\phi (n) \le \phi (N)$ for whatever $N >2024$

I thought to use prime factorisation of $N$ and $2024$ but it wasn't successful.

Thinking on the side of the answer being true , I thought to use $\phi (n)\ge \sqrt {n}$ for $n\ne 2,6$ but in vain.

Hints Please.

Bill Dubuque
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user-492177
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  • What is the source of this problem? Problems referencing the current year frequently appear in contests and such. – lulu Sep 13 '24 at 20:35
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    Hint: Bound $n \prod \frac{p-1}{p}$. $\quad$ Good candidates look like $ \prod p_i$. – Calvin Lin Sep 13 '24 at 20:36
  • Phd Entrance Exam of Indian Statistical Institute – user-492177 Sep 13 '24 at 20:36
  • Is this an ongoing exam? – lulu Sep 13 '24 at 20:40
  • Note that, unless you miswrote the question, you aren't asked to find such an $N$ even if one (or more) exist. That makes it a whole lot easier. – lulu Sep 13 '24 at 20:41
  • @lulu No. It's already done . – user-492177 Sep 13 '24 at 20:41
  • So then ask yourself: For a fixed $m\in \mathbb N$, can there be infinitely many solutions to $\varphi(n)=m$? – lulu Sep 13 '24 at 20:42
  • @lulu I don't think so $\phi (n)=m$ for infinitely many $n$ .Though I am not sure of the proof. – user-492177 Sep 13 '24 at 20:45
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    Well, start there. The proof is not difficult. Indeed, try to prove something seemingly stronger: there are only finitely many $n$ for which $\varphi(n)≤m$. Can you see how to apply that to your problem? – lulu Sep 13 '24 at 20:47
  • @lulu Ok so if $N$ is the maximum value of $n$ such that $\phi ( n)=m$ .Then $\forall n>N$ we must have $\phi(n) \ne m$ . Then sir ? – user-492177 Sep 13 '24 at 20:53
  • @CalvinLin can you elaborate a bit more sir ? – user-492177 Sep 13 '24 at 20:55
  • That's true, but it's more useful to say"if $N$ is the maximum $n$ such that $\varphi(n)≤m$ then $\forall n>N$ we must have $\varphi(n)>m$". As another hint: pretty sure the comment from @CalvinLin points you to the least $N$ which works. – lulu Sep 13 '24 at 20:55
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    Lulu's hint is a good start. Given that you're preparing for a PhD exam, I'd strongly encourage you to think about the implications of the hints and what guidance they provide, instead of just immediately demanding more next steps. – Calvin Lin Sep 13 '24 at 20:57
  • @WillJagy Indeed (this was also the content of CalvinLin's comment to the OP, though the link is far more detailed, of course). – lulu Sep 13 '24 at 20:58
  • Let $m\ge 2025$ . For each $ 1\le i \le m$ , let $k_i=\max { n | \phi (n) = i } $ which exist by bounded inverse solutions of $\phi $. Now let $N=\max { k_i | 1 \le i \le m} $ . Thus for $1\le n \le N$ , we have $ \phi (n) \le m$ which gives $\forall n>N \implies \phi (n)> m> \phi (m)$ .But this $m>N $ is not necessarily true. I am stuck at this . @lulu – user-492177 Sep 13 '24 at 22:19
  • @CalvinLin The $N$ is dependent on the $m>2024$ I choose in above comment .So how can I minimise or give an upper bound as suggested by you so that $m>N$ – user-492177 Sep 13 '24 at 22:32
  • You already mentioned a bound of the form $\varphi(n)≥\sqrt {\frac n2}$.... (See, e.g., this question for a proof). – lulu Sep 13 '24 at 22:39
  • Oh, you wanted $\varphi(n)≥\sqrt n$. For that one, see this question It really doesn't matter which one you take, either one gets the job done. – lulu Sep 13 '24 at 22:42

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