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I was studying Blind RSA signature scheme (wiki) where we have a standard RSA setup:

  • $p, q$ are some prime numbers
  • $n = pq$
  • $ed = 1\; \text{mod }\varPhi(n) $

On top of that, there is some random number $r \in Z_n$

In the Wikipedia page it says that a mapping $r \rightarrow r^e\;\text{mod } n$ is a permutation of $\{0,\dotsc,n-1\}$

Could someone explain to me why this is a case? I was trying to figure it out on my own, but I haven't done modular arithmetics in a while. I suspect that in general every mapping $r \rightarrow r^x \;\text{mod } n$, where $x \in Z^*_{\varPhi(n)}$ is a permutation of $\{0,\dotsc,n-1\}$ but I have no clue where it comes from.

kopi22
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  • Can you show that it is injective and surjective? Hint: If $ r^e = s^e$, raise both sides to the power $d$. – Calvin Lin Nov 08 '21 at 19:45
  • Maybe something here might help. – Matcha Latte Nov 08 '21 at 19:49
  • @CalvinLin

    I forgot I could just show that that $r \rightarrow r^e$ is a bijection. So for the injective property we have: for every $r,s \in Z_n$ if $r^e = s^e;\text{mod }n$, then $(r^e)^d = (s^e)^d ;\text{mod }n$ hence $r = s$ so it is injective. For surjective property: for every $s \in Z_n$ there is $r=s^d$ such that $r^e=(s^d)^e = s$. Hence it is a permuation on ${0,\dotsc,n-1}$.

     – kopi22 Nov 08 '21 at 20:08
  • The answers in the linked dupe should answer your question. If not, then please elaborate. – Bill Dubuque Nov 08 '21 at 20:38

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