How to prove congruence of a number to a power

Question

How can I prove that

$$\color{blue}{ m^{abcd} = m \mod pq,} $$

for which $p$ and $q$ are prime numbers and the following equations are satisfied

$\tag{1}ab ≡ 1 \mod (p-1)(q-1).$ $\tag{2} cd ≡ 1 \mod (p-1)(q-1).$

Thanks in advance for your help!

Answer 1

From Euler's Theorem and from the multiplicativity of the Euler's totient function we get that \begin{align*} m^{\phi(pq)}&=1 \mod pq \\ m^{\phi(p)\phi(q)}&=1 \mod pq \\ m^{(p-1)(q-1)}&=1 \mod pq \\ m^{pq-p-q+1}&=1 \mod pq. \end{align*} We know that $abcd=1 \mod (p-1)(q-1)$ because of the hypothesis, so $$abcd-1=k(pq-p-q+1) \quad \text{for }k \in \Bbb Z.$$ So, we get that $$m^{abcd}=m\cdot (m^{pq-p-q+1})^k=m \cdot 1^k= m \mod pq.$$

Answer 2

${\rm mod}\ \phi:\,\ ab\equiv 1\equiv cd\,\Rightarrow\,ab(cd)\equiv 1\,$ $\Rightarrow$ $\,\color{#0a0}{abcd = 1+n\phi},\,$ therefore

${\rm mod}\,\ pq\!:\,\ m^{\large\color{#0a0}{ abcd}}\equiv m^{\large\color{#0a0}{1+n\phi}}\equiv m(\color{#c00}{m^{\large \phi}})^{\large n}\equiv m(\color{#c00}1)^{\large n}\equiv m\ \ {\rm by}\ \ \color{#c00}{m^{\large \phi}\equiv 1}\ \ \text{(Euler)}$

Remark $\ $ Generally if $\ m^{\large k}\equiv 1\,$ then $\ i\equiv j\pmod{\!k}\,\Rightarrow\, m^{\large i}\equiv m^{\large j}.\, $ In other words, if we know that $\,m^{\large k}\equiv 1\,$ then it is valid to consider all exponents on $\,m\,$ to be modulo $\,k\,$ (proof as above).