Prove congruence for all integers

Question

Prove that for all $a\in \mathbb{Z}$, following holds $$a^{145}\equiv a\ \ \ (\text{mod }133)$$

I could use some hints on how to approach this problem.

Answer 1

$133=19\cdot7$

For any prime $p$ either $p|a$ or $p\nmid a\iff(p,a)=1$

For $19,$ either $19|a\implies19|(a^{145}-a)$

or $(a,19)=1$ By Fermat's Little Theorem, $a^{19-1}\equiv1\pmod{19}$

$a^{145}-a=a[(a^{18})^8-1]\equiv0\pmod{19}$

Similarly for $7$

Finally lcm$(19,7)$ divides $(a^{145}-a)$

In fact, using Carmichael function $\lambda(133)=18,$

we can prove $a^{18n+1}\equiv a\pmod{133}$ where integer $n\ge0$

Answer 2

Hint $\ $ Appply the following slight generalization of $\color{#c00}{\rm Fermat's}$ little Theorem: for $\,p=7,19$

• $ $ If $\ p\!-\!1\mid e\!-\!1\,$ then $\, {\rm mod}\ p\!:\ a^e\equiv a.\,$ Indeed, it is true if $\,a\equiv 0,\,$ and if $\,a\not\equiv 0\,$ then $\,e\, =\, 1+(p\!-\!1)k\ $ so $\ a^e\equiv a (\color{#c00}{a^{p-1}})^k\overset{\rm\color{#c00}{ Fermat}}\equiv a \color{#c00}{(1)}^k\equiv a,\ $ so $\ a^e\!-a\equiv a-a\equiv 0.$

Remark $\ $ Above we used basic congruence rules. The key idea generalizes, see Korselt's Carmichael Number Criterion.