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Question: Prove that for any given $n \in \mathbb N$, $\phi(x) = n$ has only finitely many solutions.

My attempt:

Let there be infinite solutions for

$$\phi(x) = n$$

so every natural number greater than $x$ has only $n$ primes before it. Thus, every number after the last prime is composite.

Now, let the finite set of primes be $p_1,p_2,...,p_n$.

Now, the number $p_1p_2...p_n + 1$ is relatively prime to all the finite primes, thus it is a new prime

This contradiction violates our assumption of a finite set of primes, so our assumption of infinite solutions is wrong.

Hence, $\phi(x) = n$ has only finite solutions.

Issue: I feel that the approach of $p_1p_2...p_n + 1$ is wrong and doesn't prove that the number itself is prime, but since it is of form $1 \text{ mod } p_i$ for $i \epsilon \text{ 1,2,...,n }$, thus it must be prime to all of them. Also, something about this proof feels off in terms of the assumptions. Any help is very appreciated!

PS: I know this is a duplicate but since I want to know what is wrong with my approach, I thought of sharing my method to get the explanation over the finite set of prime logic and the proof approach

calculatormathematical
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Ayush Maurya
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  • "every natural number greater than $x$ has only $n$ primes before it". What does this mean? $x$ is not fixed, it takes infinitely many values. – ultralegend5385 Mar 30 '24 at 07:09
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    There's nothing wrong concluding that $p_1p_2..p_n + 1$ is a prime because you started from a false assumption, and from falseness everything follows (explosion principle). The important thing is that you reach a contradiction withou additional assumption – ioveri Mar 30 '24 at 09:13
  • You haven't defined what the function $\phi(x)$ is. If you mean it to be the number of primes up to $x$, then $\pi(x)$ is the more common notation ($\phi$ is usually the Euler totient function). – Greg Martin Mar 30 '24 at 09:33
  • We already know that there are infinitely many primes, so there's no need to go through Euclid's argument again. Since there are infinitely many primes, $p_{n+1}$ exists, and by definition $\pi(x) \ge n+1$ for all $x\ge p_{n+1}$. So there are at most $p_{n+1}-1$ solutions to $\pi(x) = n$. The proof doesn't need to be any longer than that. – Greg Martin Mar 30 '24 at 09:34

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Edit: new answer due to change in OP question

Suppose $x$ has the following factorization: $$x = \prod_{p|n}p^{k_p}$$ Using the Euler's product formula, we have $$\phi(x) = \prod_{p|n}p^{k_p-1}(p-1)$$ Therefore $$\frac{\phi^2(x)}{x} = \prod_{p|n}p^{k_p-2}(p-1)^2 \geq \prod_{p|n}\frac{(p-1)^2}{p}$$

since $k_p \geq 1$. Notice that $\dfrac{(p-1)^2}{p} > 1$ for $p >2$, therefore: $$\frac{\phi^2(x)}{x} \geq \frac{1}{2} \Rightarrow \phi(x) \geq \sqrt{\frac{x}{2}}$$

If $\phi(x)=n$ then $n\geq\sqrt{\dfrac{x}{2}}$ therefore $x\leq2n^2$. Thus there can only be a finite number of solutions to $\phi(x) = n$

ioveri
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  • Did you assume $\phi $ is increasing? – calc ll Mar 30 '24 at 07:45
  • @insomesense $\phi(x)$ in this context is "the number of primes not greater than $x$" so yes it is increasing. – ioveri Mar 30 '24 at 09:09
  • $\phi$ is Euler's totient function. $\phi(5)=4, \phi(6)=2.$ So it's not increasing. It sounds like you may be thinking of the prime counting function. – calc ll Mar 30 '24 at 09:37
  • @insomesense "so every natural number greater than x has only n primes before it". This is what OP said so I assume $\phi(x)$ is the prime counting function in this context. – ioveri Mar 30 '24 at 09:51
  • Then it's terrible notation. Especially considering the tag: totient function. – calc ll Mar 30 '24 at 10:02
  • @insomesense oh I didn't even notice the tag. Yeah I think OP understood the notation wrongly. – ioveri Mar 30 '24 at 10:07
  • Sorry to all because what I had in mind was Euler's totient function. I forgot that the function counts the relatively prime numbers and not the primes before it. I am struggling. – Ayush Maurya Mar 30 '24 at 15:13
  • @AyushMaurya please edit your question accordingly, I also modified my answer. – ioveri Mar 30 '24 at 15:43
  • there is a bunch of stuff at https://math.stackexchange.com/questions/301837/is-the-euler-phi-function-bounded-below including the Nicolas criterion for RH – Will Jagy Mar 30 '24 at 18:41