The question:
The number $561$ factors as $3 \cdot 11 \cdot 17$. First use Fermat's little theorem to prove that $$a^{561} \equiv a \pmod 3 \\ a^{561} \equiv a\pmod {11} \\ a^{561} \equiv a\pmod {17}$$ for every value of $a$. Then explain why these three congruences imply that $a^{561} \equiv a (\mod 561)$ for every value of $a$.
My attempt: $$ a^2 = \left\{ \begin{array}{c} 1 (\mod 3) \quad \text{if} \quad 3 \mid a\\ 0 (\mod 3) \quad \text{if} \quad 3 \nmid a\\ \end{array} \right. \\[3ex] a^{10} = \left\{ \begin{array}{c} 1 (\mod 11) \quad \text{if} \quad 11 \mid a\\ 0 (\mod 11) \quad \text{if} \quad 11 \nmid a\\ \end{array} \right. \\[3ex] a^{16} = \left\{ \begin{array}{c} 1 (\mod 17) \quad \text{if} \quad 17 \mid a\\ 0 (\mod 17) \quad \text{if} \quad 17 \nmid a\\ \end{array} \right. $$ I'm really not sure where to go from here. The fact that $561 = 3\cdot 11 \cdot 17$ must fit in somehow, but beyond that I don't know.
Presumably you can see that $$a^{k} \equiv \left\{ \begin{array}{c} 1 \bmod p \quad \text{ if } p \nmid a\\ 0 \bmod p \quad \text{ if } p \mid a\\ \end{array} \right. $$
immediately gives $a^{k+1} \equiv a \bmod p$ and indeed $a^{nk+1} \equiv a \bmod p$
The key next step is to examine the factors of $561-1=560$.
$560 = 2^4\cdot5\cdot7$
And in particular, note
$\begin{align} 2 &\mid 560 \\ 10 &\mid 560 \\ 16 &\mid 560\end{align}$
Once you have demonstrated the three asserted equivalences to the individual primes, the result for the composite value follows immediately from "simple" equal values in the Chinese Remainder Theorem: given $b,c,$ coprime:
$\left .\begin{align}x\equiv a \bmod b \\x\equiv a \bmod c \end{align}\right\}\implies x\equiv a \bmod bc$
Of interest: $561$ is the smallest Carmichael number