0
$\begingroup$

Has the congruence $n\equiv\varphi(n) \pmod p$, with $p$ being an odd prime not dividing $n$, been examined before?

Because it is easy to find solutions for $n$ with few primes in its decomposition (e.g., $n=14007$, $p=5$), a classification of solutions with respect to $\omega(n)$ would be interesting.

$\endgroup$
5
  • $\begingroup$ I suppose that you are really thinking about primes which do not divide $n$, since the congruence holds whenever $p^{2}$ divides $n$ with $p$ prime. $\endgroup$ Aug 17, 2016 at 18:28
  • 3
    $\begingroup$ Actually, it is in some sense routine to calculate the primes $p$ which divide $n - \phi(n)$: Let $n = p_{1}^{m_{1}} \ldots p_{r}^{m_{r}}.$ Then the prime divisors of $n - \phi(n)$ are the primes $p_{i}$ with $m_{i}>1$, together with the primes $p$ which divide $\prod_{i=1}^{r} p_{i} - \prod_{i=1}^{r}(p_{i}-1).$ $\endgroup$ Aug 17, 2016 at 18:36
  • $\begingroup$ I have edited my question $\endgroup$ Aug 18, 2016 at 6:42
  • $\begingroup$ $n-\varphi(n)$ in general has additional different prime Divisors than $n$ $\endgroup$ Aug 18, 2016 at 6:45
  • 1
    $\begingroup$ Yes indeed, as my second comment makes clear. With the notation as in that comment, you are looking at the prime divisors of $\sum_{ \phi \neq J \subseteq I} (-1)^{|J|} \prod_{ i \in I \backslash J} p_{i}$, where $I = \{1,2,\ldots ,r\}$, together with the primes which divide $n$ to thhe second or higher power. $\endgroup$ Aug 18, 2016 at 8:36

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.