Given that $a \equiv b \pmod n$ and $c\equiv d \pmod n$, I need to prove that $ac \equiv bd \pmod n$
So far, I've only managed to deduce that $a+b \equiv c+d \pmod n$. I don't know if this is usable, but it's there, at least.
Any hint would be greatly appreciated!
Hint:
$$ac-bd=ac-ad+ad-bd=a(c-d)+d(a-b)=\;\ldots$$
Like Gauss did in his Disquisitiones Arithmeticae: $$ a\equiv b\pmod{n} $$ implies $$ ak\equiv bk\pmod{n} $$ for all $k$.
Suppose now $a\equiv b\pmod{n}$ and $c\equiv d\pmod{n}$. Then $$ ac\equiv bc\pmod{n} $$ and $$ bc\equiv bd\pmod{n}. $$ By transitivity, $$ ac\equiv bd\pmod{n}. $$
a ≡ b (mod n) means a = b + xn for some x. Work with that definition.
$\begin{align} {\bf Hint}\ \ \ a\ &=\ \color{#c00}b\ \ +\,n\,j\\ c\, &= \ \ \color{#0a0}d\,+\,n\,k\\ \Longrightarrow\,\ ac &= \color{#c00}b\color{#0a0}d\,+\,n\:\!(\ldots)\ \ \text{for an } {\it integer} \ (\ldots) \end{align}$
Remark $\, $ If $\,n = 10\,$ then this generalizes a units digit rule well-known from decimal arithmetic, namely mod $10,\,$ the units digit of a product is congruent to the product of the unit digits, e.g. $\,1\color{#c00}3\cdot 1\color{#0a0}6 = 208\,$ has units digit $\,\color{#c00}3\cdot\color{#0a0}6\equiv 8.\,$ Said in the language of the Congruence Product Rule
$$\begin{eqnarray}{\rm mod}\ 10\!:\ &&1\color{#00}{3}\equiv \color{#c00}3,\ \ 1\color{0a0}{6}\equiv \color{#0a0}6\\[.2em] \Longrightarrow\ &&1\color{c00}{3}\cdot 1\color{0a0}{6}\equiv \color{#c00}3\cdot \color{#0a0}6\equiv 8\end{eqnarray}\qquad\qquad\qquad $$
We can view the Congruence Product Rule as a radix $\,n\,$ generalization of the units digit product rule. However, it is more general, since the "units digits" $\,\color{#c00}b,\color{#0a0}d\,$ need not lie in the interval $\,[0,n\!-\!1].$
The prior linked post has proofs of all of the common congruence arithmetic rules.
