modular arithmetic - $a equiv b pmod n$ and $cequiv d pmod n$ implies $ac equiv bd pmod n$

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!

Answer

$\begin{align} {\bf Hint}\ \ \ a\, &=\ b\ \ +\,n\,j\\ c\, &= \ \ d\,+\,n\,k\\ \Rightarrow\,\ ac &= bd\,+\,n(\_\_)\ \ \text{for an integer }\, (\ldots) \end{align}$

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

$$\begin{eqnarray}{\rm mod}\ 10\!:\ &&1\color{#00}{3}\equiv \color{#00}3,\ \ 1\color{0a0}{6}\equiv \color{0a0}6\\ \Rightarrow\ &&1\color{c00}{3}\cdot 1\color{0a0}{6}\equiv 3\cdot 6\equiv 8\end{eqnarray}\qquad\qquad\qquad$$

The Congruence Product Rule may be viewed as a radix $\,n\,$ generalization of the units digit product rule. However, it is more general, since the "units digits" $\,b,d\,$ need not lie in the interval $\,[0,n\!-\!1].$

See here for proofs of all of the common congruence arithmetic rules.