Is $(((a \text{ mod } n)* a \text{ mod } n) * a \text{ mod } n) = a*a*a \text{ mod } n$?

Question

The question is if the modulus of a multiplication, i.e. $a*a*a$ modulus $n$, is the same when we take the modules at each step of the multiplication. So if

$(((a \text{ mod } n)* a \text{ mod } n) * a \text{ mod } n) = a*a*a \text{ mod } n$?

Answer 1

Yes, because $\mathbb{Z}\rightarrow\mathbb{Z}/(n),a\mapsto a+(n)$ is a homomorphism, the so called canonical homomorphism. To see this You just have to prove that multiplication on $\mathbb{Z}/(n)$ is well defined: Indeed if $b-b'\in(n)$ then $$ab'-ab=a(b'-b)\in (n).$$ Thus $(a+(n))(b+(n))=(a\mod(n))(b\mod(n))=ab+(n)=ab\mod(n)$ is well defined.

Answer 2

If $a\equiv b(\mod n)$ and $a'\equiv b'(\mod n)$, then $aa'\equiv bb' (\mod n). $ So you can either multiply and then reduce $\mod n$ or vice versa. For an application, to see that you can reduce large computations to something with logarithmic runtime, e.g. have a look at square & multiply