I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things:
A composite number $m$ is called a Carmichael number if the congruence $a^{m-1} \equiv 1 \pmod{m}$ is true for every number a with $\gcd(a,m) = 1$.
Verify that $m = 561 = 3 \times 11 \times 17$ is a Carmichael number.
Solution given:
Apply Fermat's Little Theorem to each prime divisor of $m$: \begin{align*} a^2 &\equiv 1 \pmod{3}\\ a^{10} &\equiv 1 \pmod{11}\\ a^{16} &\equiv 1 \pmod{17} \end{align*} This somehow then implies that $a^{80} \equiv 1 \pmod{561}$ then accordingly $a^{560} \equiv 1 \pmod{561}$.
I am lost as to how the 3 congruences imply $a^{80} \equiv 1 \pmod{561}$ ($80 = \mathrm{LCM}(2,10,16)$).
Can somebody clarify this for me?
Thanks!