7

Is it true that for every $k \in \mathbb{N}$ there exists a natural number $x$ such that $\phi(x)=\phi(x+1)=\cdots=\phi(x+k)$, where $\phi$ is the Euler's totient function?

I thought about this problem today and I've no idea how to solve it or even attack it.

Also, I would be interested in variants of this problem, in which we substitue $\phi$ for another number-theoretical interesting functions, like $\sigma,\tau,\lambda$, etc.

Thank you in advance!

u1571372
  • 3,205
  • I assume you mean $k$ consecutive naturals? – user217285 Jul 08 '15 at 19:28
  • 1
    For what it's worth: to the best of my knowledge this is unknown. Looks hard. Mathworld tells us that the only 3 consecutive integers under $10^{10}$ with the same totient value are 5186,5187,5188. http://mathworld.wolfram.com/TotientFunction.html – lulu Jul 08 '15 at 19:28
  • @Nitin Oh, I assumed the OP wanted consecutive values. Upon rereading, I see that isn't stated. Might make it a lot easier if you drop that... – lulu Jul 08 '15 at 19:30
  • Sorry, I've edited the question. I meant consecutive naturals. – u1571372 Jul 08 '15 at 19:31
  • No problem. The other question might be interesting too, and looks a lot more tractable. But I still don't know how to do it. – lulu Jul 08 '15 at 19:36
  • For the consecutive integers one, there's a good chance it is false. See: https://en.wikipedia.org/wiki/Euler%27s_totient_function#Ratio_of_consecutive_values. – user217285 Jul 08 '15 at 19:39
  • @Nitin: I'm not sure how much those results bear on the present question? Certainly we'd expect $\phi$ to vary a lot, and those results quantify that, but they don't make it seem yet less likely to me that such $x$ might exist for all $k$ than it already did anyway. – joriki Jul 08 '15 at 20:42
  • 8
    Erdös conjectured in 1945 that the answer is yes: see http://www.ams.org/journals/bull/1945-51-08/S0002-9904-1945-08390-6/ DOI: http://dx.doi.org/10.1090/S0002-9904-1945-08390-6 – Robert Israel Jul 08 '15 at 22:17
  • Very hard question... I tend to think that is false from the definition of totient $\varphi(n) =n \prod_{p\mid n} \left(1-\frac{1}{p}\right)$. – Masacroso Jul 08 '15 at 22:32

0 Answers0