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Let's consider $a_1$ and $a_2$: \begin{align*} a_1&=a(p)\\ a_2&=a(q) \end{align*} $r_1$ and $r_2$ are the orders of $a_1$ and $a_2$ in the rings $\mathbb Z_p$ and $\mathbb Z_q$. $$a_1^r(p) = a^r(p)=1,$$ because $a^r=1(pq)$. Since $r_1$ is the order of $a_1$ in $\mathbb Z_p$, the number $r$ is a multiple of $r_1$. The same reasoning for $\mathbb Z_q$ gives us $r$ is a multiple of $r_2$.

Definition of variables

  1. $N = p \cdot q $, where $p$ and $q$ are primes

  2. $gcd(a,N) = 1$ that is $a$ and $N$ are co-primes.

  3. Periodic function is described as$ Fa(x) = a^x mod N $

  4. $a^r \equiv 1 \pmod N$ that is $r$ is the period of the Fa(x)
    Now my question is How the sixth line, How is it coming?

How does $\,a^r \equiv 1\pmod{\!pq}\,$ lead to $\,a_1^r (p) = a^r(p) = 1$?

tryst with freedom
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Saptarshi Sahoo
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1 Answers1

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I cant understand how $\,a^r \equiv 1\pmod{pq}$ is leading to $\,a_1^r (p) = a^r(p) = 1$

Recall that congruences persist mod $\,\rm\color{#c00}{factors}\,$ of the modulus.

So $\!\bmod \color{#c00}pq\!:\ 1\equiv a^{\large r}\Rightarrow\bmod \color{#c00}p\!:\ 1\equiv a^{\large r}\equiv a_1^{\large r}\ $ by $\,a\equiv a_1\,$ and the Congruence Power Rule.

Hence since $\,a_1\,$ has order $\,r_1\,$ we infer $\, r_1\mid r\,$ by the Order Theorem.

Similarly $\, r_2\mid r,\ $ hence $\,{\rm lcm}(r_1,r_2)\mid r\,$ by the LCM Universal Property,

Furthermore, recall that $\,{\rm lcm}(r_1,r_2) = r_1 r_2\,$ when $\,\gcd(r_1,r_2) = 1$.

Bill Dubuque
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  • what does r1 | r mean? and Could you please give some more description on the phrase "Congruences persist mod factors of the modulus". I cant quite get my head around it. – Saptarshi Sahoo Dec 28 '19 at 13:07
  • @Uzumaki $\ a\mid b,$ means $a$ divides $b,,$ i.e. $,an = b,$ for some integer $n.\ $ Please see the linked post for a proof of said persistence. – Bill Dubuque Dec 28 '19 at 21:09