Verifying Carmichael numbers

Solution 1:

Note that $80 = \mathrm{lcm}(2,10,16)$. So you can write $a^{80} = (a^2)^{40} = (a^{10})^{8} = (a^{16})^5$. So, \begin{align*} a^{80}= (a^2)^{40} &\equiv 1^{40} = 1\pmod{3},\\ a^{80}= (a^{10})^{8} &\equiv 1^8 = 1 \pmod{11},\\ a^{80}= (a^{16})^5 &\equiv 1^5 = 1\pmod{17}. \end{align*}

By the Chinese Remainder Theorem, the system of congruences \begin{align*} x&\equiv 1\pmod{3}\\ x&\equiv 1\pmod{11}\\ x&\equiv 1\pmod{17} \end{align*} has a unique solution modulo $3\times 11\times 17 = 561$. But both $x=1$ and $x=a^{80}$ are solutions. Since the solution is unique modulo $561$, then the two solutions we found must be congruent. That is, $$a^{80}\equiv 1\pmod{561}.$$

(Added. Or, more simply, as Andres points out, since $3$, $11$, and $17$ each divide $a^{80}-1$, and are pairwise relatively prime, then their product divides $a^{80}-1$).

Once you have that $a^{80}\equiv 1\pmod{561}$, then any power of $a^{80}$ is also congruent to $1$ modulo $561$. In particular, $$a^{560} = (a^{80})^{7} \equiv 1^7 = 1 \pmod{561}$$ as desired.

Solution 2:

HINT $\: $ For primes $\rm\ p\neq q\:$ coprime to $\rm\:a\:,\:$ if $\rm\ p-1,q-1\ |\ m\ $ then $\rm\ p,q\ |\ a^m - 1\ \Rightarrow\ pq\ |\ a^m - 1$ since lcm = product for coprime integers (here distinct primes).