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As my most recent question still does not have any answers and it appears to be a difficult problem, I propose the following problem (that seems easier), but which I still could not manage to solve:

Is it true that for every $k \in \mathbb{N}$ there exists distinct natural numbers $x_1, \cdots, x_k$ such that $\phi(x_1)=\phi(x_2)=\cdots=\phi(x_k)$, where $\phi$ is the Euler's totient function?

Any ideas? Thank you!

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u1571372
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    This is true, but far from trivial. See http://www.math.uiuc.edu/~ford/wwwpapers/sierp.pdf – Servaes Jul 09 '15 at 22:44
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    Ford's statement is stronger than the one stated here: vieisci is asking for at least $k$ solutions to $\phi(x) = y$ for some $y$, but Ford wants exactly $k$. – Robert Israel Jul 09 '15 at 22:48
  • @Servaes, thank you for your answer. Any hope that might exist an easier argument if one only asks for at least $k$ solutions? – u1571372 Jul 09 '15 at 22:51
  • @ElliotG, sorry, I meant distinct natural numbers, I've edited the question. – u1571372 Jul 09 '15 at 22:51
  • $\phi(p)=2$ for every prime number $p$. And $\phi(2n)=\phi(n)$ for $\gcd(2,n)=1$ where $n \in \mathbb{N}$ – Lucky Chouhan Dec 05 '23 at 07:11

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Let $N(x)$ be the number of different values of $\phi(n)$ that are $\le x$. Pillai (Bull. Amer. Math. Soc. Volume 35, Number 6 (1929), 832-836) showed that $N(x) = O(x/(\log x)^t)$ where $t = \log(2)/e$. In particular, $N(x)/x \to 0$ as $x \to \infty$. If you take $x$ large enough that $N(x)/x < 1/k$, then by the Pigeonhole Principle some $k$ natural numbers $\le x$ must have the same value of $\phi$.

Robert Israel
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