2025/001) If $GCD(a,35) = 1$ then show that $a^{12} \equiv 1 \pmod {35}$

 We have $35 = 7 * 5$ so

$GCD(a,5) = 1$ and $GCD(a,7) = 1$

Because GCD(a,5) is 1 so as per Fermats Little Thorem $a^{4} \equiv 1 \pmod {5}$ or 

$a^{12} \equiv 1 \pmod {5}\cdots(1) $

Because GCD(a,7) is 1 so as per  Fermats Little Thorem $a^{6} \equiv 1 \pmod {7}$ or

 $a^{12} \equiv 1 \pmod {7}\cdots(2)$

From (1) and 2

 $a^{12} \equiv 1 \pmod {35}$