5
$\begingroup$

Find $n\in \mathbb N$ so that $(\mathbb Z_n, +, \cdot)$ has exactly 4 invertible elements and 5 zero-divisors.

As I couldn't find any theorem that would lead me toward a solution, so I have been trying guessing and checking with no results so far, so I am asking for a push in the right direction.

  • 1
    Is it exactly 4 invertible elements and exactly 5 zero-divisors?2012-09-01
  • 0
    @lhf yep! I am sure.2012-09-01
  • 0
    In this case, $n$ would have to be 9 but $\mathbb Z_9$ has 6 units.2012-09-02
  • 0
    Also, $\phi(n)=4$ iff $n=5, 8, 10, 12$.2012-09-02
  • 1
    Perhaps we are using a definition whereby zero is not a zero-divisor.2012-09-02

1 Answers 1