On $\sup|\varphi^{-1}(n)|=+\infty$

Question

I am trying to find an elementary proof of the following fact:

Given some $N\geq 2$, there are $N$ distinct integers $a_1,\ldots,a_N$ such that $\varphi(a_1)=\ldots=\varphi(a_N)$ with $\varphi$ being Euler's totient function.

My original analytic proof goes as follows: If the number of solutions of $\varphi(x)=N$ were bounded, the series $$ \sum_{n\geq 2019}\frac{1}{\varphi(n)\log^2\varphi(n)}$$ would be convergent by comparison with $\sum_{n\geq n_0}\frac{1}{n \log^2 n}$ and condensation. It is enough to show that the last series is divergent. It is bounded below by a multiple of $$ \sum_{n\geq 2019}\frac{\sigma(n)}{n^2 \log^2(n)}$$ and since $$ \sum_{n\geq 1}\frac{\sigma(n)}{n^{2+s}}=\zeta(s+1)\zeta(s+2)$$ for any $s>0$, it is enough to prove that the integral $$ \int_{0}^{+\infty}\int_{0}^{+\infty}\zeta(1+t+s)\zeta(2+t+s)\,ds\,dt $$ is divergent, or that the integral $$ \int_{0}^{+\infty} u\,\zeta(1+u)\zeta(2+u)\,du $$ is divergent. On the other hand this is trivial since $u\zeta(1+u)\zeta(2+u)\geq u$ for any $u>0$.

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Alternative combinatorial proof: we may consider a very large $N$ and the numbers in $[N,2N]$ with at least $\log\log\log N$ prime factors. They have a positive density in $[N,2N]$, and they are mapped by the totient function into an interval with length $O\left(\frac{N}{\log \log N}\right)$. By the pigeonhole principle, at least $\Omega(\log\log N)$ elements of $[N,2N]$ share the same $\varphi$.

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I would be happier in having a combinatorial proof possibly not relying on subtle statements about the average order of $\omega(n)$ or Mertens' theorem about $\sum_{p\leq x}\frac{1}{p}$.

Answer 1

Here it is an elementary argument inspired by Pillai$^{(*)}$: the point is that $\nu_2(\varphi(n))\approx\omega(n).$

Let $O^{\omega}$ the set of odd natural numbers with at least $\omega$ prime divisors and let $O^{\omega}_n=O^{\omega}\cap[1,n]$.
$O^{\omega}$ certainly has density $\frac{1}{2}$ for any $\omega$ - it is enough to invoke some elementary version of the PNT and basic sieving arguments. The totient function maps $O^{\omega}_n$ into a subset of $[1,n]\cap 2^{\omega}\mathbb{N}$, hence by the pigeonhole principle at least $(1-\varepsilon_n)2^{\omega-1}$ elements of $O^{\omega}_n$ share the same $\varphi$. Since $\omega$ is arbitrary we have finished.

^{His surname is pure magic: Sivasankaranarayana.}