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I recently attempted to solve the following question (exercise 0.2.10 from Dummit & Foote - Abstract Algebra):

Prove for any given positive integer $N$ there exist only finitely many integers $n$ with $\varphi(n) = N$, where $\varphi$ denotes Euler’s $\varphi$-function. Conclude in particular that $\varphi(n)$ tends to infinity as $n$ tends to infinity.

I ended with something much similar to:

  1. Is there an integer $N>0$ such that $\varphi(n) = N$ has infinitely many solutions?
  2. Prove for any given positive integer $N$ there exist only finitely many integers n with $φ(n)=N$

Which show that, for any given $N$, there are finitely many $n$ such that $\varphi(n) = N$. I am OK with this. However, the question also asks whether $\varphi(n)$ tends to infinite as $n$ does; 2. attempts to conclude that from 1, and that doesn't feel right to me. Let me explain:

All answers reach a point where for $\varphi(n) = N$, then $n$ must be a product of finitely many bounded primes, each raised to a bounded exponent. However, nothing is said about a lower bound on such $n$, and thus it seems wrong to conclude that $N$ grows with $n$; what if $N$ oscillates without a lower bound as $n$ grows bigger? The argument doesn't seem to address this possibility.

I know that $\varphi(n)$ is in fact bounded below, and many good answers can be found here: Is the Euler phi function bounded below?

What I ask, then, is whether that conclusion is false, needs some tinkering to work, or is fine and I missed something.

Alek Fröhlich
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  • $\varphi(n)$ is a function on the natural numbers so it can be viewed as a sequence. What is the definition of $\lim_{n \rightarrow \infty}\varphi_n = \infty$? Can you prove the statement using the definition directly and the fact that there are only finitely many $n$ such that $\varphi(n) = N$ for any $N$? – MBW Jun 07 '20 at 02:07

1 Answers1

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The conclusion is fine. From the observation that the solution set of $\varphi(x) = N$ is finite for each fixed $N$, it follows that $\varphi$ does not "oscillate" like that. The proof follows, so skip the rest if you'd like to work it out yourself!

Recall that the sequence $\varphi$ tends to infinity precisely if for any $M \in \mathbb{R}$ we can find an integer $K$ such that for all $n > K$, $a_n > M$.

So take any number $M$. There are finitely many integers $i$ that lie between $0$ and $M$. We know by our observation that for each of these finitely many integers $i$ the solution set of the equation $\varphi(x) = i$ is itself finite. But the union of finitely many finite sets is still finite, so there are finitely many solutions to $\varphi(i) \leq M$. The finite set of solutions has a maximal element, denote it $K$. Now for all $n > K$, $n$ is not a solution to $\varphi(n) \leq M$, so in fact we have $\varphi(n) > M$.

Z. A. K.
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