<s>[INST] Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63}$. Why is this assumption necessary? Question:
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63} $. Why is this assumption necessary?

Proof:
Since $\gcd(a,3)=1$ $\Leftrightarrow a\equiv 1\pmod 3$ $\Leftrightarrow a^7\equiv 1\pmod3\equiv a\pmod3$
Then using Fermat's Little Theorem:
If $a,p\in\mathbb N$ and $p$ is prime then $a^7\equiv a\pmod7$
$\Rightarrow 3 |a^7-a$ and $7 |a^7-a$
$\Leftrightarrow a^7-a=3k_1$ and $a^7-a=7k_2$
$\Rightarrow (a^7-a)^3=63(k_1)^2k_2$ $\Rightarrow (a^7-a)^3\pmod{63}\equiv 0$
Since $x^m\pmod n\equiv x\pmod n$
$\Rightarrow (a^7-a)^3\pmod{63}\equiv a^7-a\pmod{63}\equiv 0$
$\Leftrightarrow a^7\equiv a\pmod{63}$
The thing I am struggling with is that the question says that the only assumption necessary is that $\gcd(a,3)=1$. Surely there are two assumptions neccessary, since to use Fermat's Little Theorem (in this situation) we need $a\neq 0 \pmod7 \space\space\space(\gcd(a,7)=1)$. I am sure there is something obvious I'm missing -would be great if someone could check over what I have done and point out any mistakes :)

 [/INST] $\gcd(a,7) = 1$ is not necessary. If $\gcd(a,7)=7$ then $a^7\equiv a \equiv 0 \pmod 7$ and no theorem needed.
FLT says if $\gcd(a,7)=1$ then $a^6\equiv 1\pmod 7$ and from that we conclude that $a^7 \equiv a \pmod 7$ ALWAY; trivialy so if $7|a$--- and by FLT if $\gcd(a,7)=1$.
But we need $\gcd(a,3) = 1$.
Note: $3^7 \equiv 45 \pmod {63}$ so that is a failure.
Fermat's Little theorem applies to primes; and $63= 7*3^2$ is not prime.  The prime factor of $7$ is to a single power so we can conlude $a^7\equiv a \pmod 7$.  
But for the factor of $9=3^2$ we use Euler's theorem to note $a^6 \equiv 1\pmod 9$ if $\gcd(a,3)=1$ and $a^7\equiv a\pmod 9$ if $\gcd(a,3)=1$. 
Now if $3|a$ but $9\not \mid a$ we do not have $a^7\equiv a \equiv 0\pmod 9$ trivially.
In fact we have $a^7\equiv 0 \pmod 9$ while $a^7\equiv a\pmod 7$ so
$a^7 \equiv a + 7m\equiv 0 + 9k\pmod {63}$ where $0 \le m < 9; 0\le k< 7$ 
So if $3\mid a$  we do not have $a^7\equiv a\pmod {63}$ unless $9\mid a$.
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BTW, If $9|a$ we do have $a^7\equiv a\pmod {63}$ and we do have $a^7\equiv a\pmod{21}$ always.....
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