TITLE: How to explain this property of totient?
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: 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.