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I am not a student of mathematics, and I have never studied number theory. I was looking at RSA recently. Euler Theorem is used here. I don't quiet understand $1(mod\quad n)$ in Euler Theorem$a^{\phi (n)}≡1\pmod n$. Are $1\pmod n$ and $1\; mod\; n$ the same ? Isn't $1\; mod\; n$ always one?

Why is the $ed=1\pmod n$ equal to $d=e^{-1}\;mod\;\phi(n)$?

Gerrie
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1 Answers1

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No. The issue is what that notation means. You're thinking of mod in terms of the operation. Rather, what this denotes is a congruence.

Specifically, if

$$a \equiv b \pmod n$$

what this notation means is that $n \mid a-b$. That is, $n$ evenly divides $a-b$. Or, even more clearly, there is some integer $k$ such that $a-b = kn$. Some examples and why they hold: $ \newcommand{\con}[4]{#1 \equiv #2 \!\!\!\! \pmod{#3} \;\;&\text{ because } \;\; #1 - #2 = #3 \cdot #4 \\} $ \begin{align*} \con 5 2 3 1 \con 9 0 3 3 \con{15}{1}{2}{7} \con{12}{2}{5}{2} \con{18}{5}{13}{1} \con{123}{0}{41}{3} \con{16}{7}{9}{1} \con{24}{3}{7}{3} \con{2}{4}{2}{(-1)} \con{5}{15}{5}{(-2)} \end{align*}

PrincessEev
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  • Thanks for your answer. According to your answer, a^ϕ(n) mod n ≡ 1 mod n. Am I right? The ed=1(modn) equal to d=e−1modϕ(n) int the question ,do you know why? – Gerrie Dec 04 '20 at 08:03
  • "a^ϕ(n) mod n ≡ 1 mod n. Am I right?" -- I suppose so, though if you intend to work on proofs and such regarding this, I think that definition might be a bit more cumbersome to work with. – PrincessEev Dec 04 '20 at 08:05
  • "he ed=1(modn) equal to d=e−1modϕ(n) int the question ,do you know why?" -- At a guess, the $e^{-1}$ bit likely refers to a multiplicative inverse. It's a bit much to get into here though at least in the manner in which I understand it. You would need to understand the notion of the groups or ring of integers modulo $n$. An overview seems to be given here. That said I don't know the details of the proof of Euler's theorem offhand, sorry, so I don't otherwise know why one equivalence implies the other. – PrincessEev Dec 04 '20 at 08:08