Trying to understand how RSA works. For the sake of simplicity I chose $3$ and $7$ as $p$ and $q$. I want to encode message $M=20$. Obviously, my $N=3\times 7=21$, I need to chose $e$ which is coprime with $(p-1)(q-1)=(3-1)\times (7-1)=12$, I might choose $e=5$.
Now whatever $e$ I chose now to be coprime with $12$, it seems when I compute $C=M^e(\mod 21)$ I always get the very same $C=M=20$ as the ciphtertext.
The same I think happens if I choose $p$ and $q$ to be $5$ and $7$.
I feel this is something trivial but I would like to understand what rule I have violated by choosing p and q being 3 and 7 or 5 and 7?