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I am running a program to search for solutions of $$\varphi(pm+1)=\varphi(pm+p+1).$$

So far, for $m=1,\ldots,327$ solutions have been found (some relatively large).

(in the body of the question, $p$ is a natural number, not necessarily prime)

I would like to conjecture that for every $m \in \mathbb N$ there exists natural number $p \in \mathbb N$ such that $\varphi(pm+1)=\varphi(pm+p+1)$

  • Is this known to be true?

  • If not known to be true, is there some evidence that totient could have (or not have) this interesting property?

Update: I stopped the program at $m=407$ since it was taking a long long time to find solution for $m=407$ (if there is any), but all $m=1,\ldots,406$ have a solution.

Update 2: Currently, the program is running the code for approx. half an hour just for $m=490$, and yes, the solutions exist for $m=1,\ldots,489$

Update 3: It seems that it is rare that for some $k \in \mathbb N$ there do not exist some $p$´s such that we do not have $\varphi(pm+k)= \varphi(pm+p+k)$ for all $m$´s, and a research of when that fails to be true seems to be one possible avenue of research.

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  • $\begingroup$ Ah, this is hardly true! :( $\endgroup$
    – user153451
    Commented Mar 30, 2020 at 12:05
  • $\begingroup$ The program is taking a long long time at m=407. $\endgroup$
    – user153451
    Commented Mar 30, 2020 at 12:11
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    $\begingroup$ Is this the same as math.stackexchange.com/questions/3601590/… $\endgroup$
    – Gerry Myerson
    Commented Mar 30, 2020 at 12:37
  • $\begingroup$ @GerryMyerson Not, but very different, there is investigation of pm+m+1 and here of pm+p+1, me and Peter are collaborating in some chatrooms. $\endgroup$
    – user153451
    Commented Mar 30, 2020 at 12:40
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    $\begingroup$ OK, sorry – still, I think it's good for each question to link to the other. $\endgroup$
    – Gerry Myerson
    Commented Mar 30, 2020 at 12:41

1 Answer 1

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It might just be a matter of randomness. Seeing no reason for $\varphi(pm+1)$ and $\varphi(p(m+1)+1)$ to be especially related, we might imagine heuristically that $\varphi(pm+1)$ and $\varphi(p(m+1)+1)$ have probability $\sim \text{constant}/(pm)$ of being equal. Since $\sum_m 1/m = \infty$, it would then be reasonable to expect there to be infinitely many $m$ for which this is the case. Of course this is not a proof.

This also suggests that if a small $m$ is not found for a particular $p$, you might need to look very far (something like $\exp(\text{constant}/p)$) before finding an $m$ that works.

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    $\begingroup$ I have found some $k$ for which for some small $m$ it seems that there is no $p$ such that $\varphi(pm+k)=\varphi(pm+p+k)$ but both such $k$ and $m$ seem to be very rare. $\endgroup$
    – user153451
    Commented Mar 31, 2020 at 6:45
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    $\begingroup$ Nice answer. Although the "matter of randomness" actually involves nothing random... $\endgroup$
    – WhatsUp
    Commented Mar 31, 2020 at 7:13
  • $\begingroup$ Agreed, it's not really random, but a random model can give useful heuristics, as with Cramér's random model for the primes. $\endgroup$
    – Robert Israel
    Commented Mar 31, 2020 at 20:34

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